SequencesandSeries

Question1

Thenumberofcommontermsintheprogressions4,9,14,19,......,upto25th

termand3,6,9,12,upto37thtermis:

[27-Jan-2024Shift1]

Options:

Answer:C

Solution:



-------------------------------------------------------------------------------------------------

Question2

[27-Jan-2024Shift1]

Answer:9

Solution:



-------------------------------------------------------------------------------------------------

Question3

The20thtermfromtheendoftheprogression

[27-Jan-2024Shift2]

Options:

-118

-110

-115

-100

Answer:C

Solution:



-------------------------------------------------------------------------------------------------

Question4

IfinaG.P.of64terms,thesumofallthetermsis7timesthesumoftheodd

termsoftheG.P,thenthecommonratiooftheG.P.isequalto

[29-Jan-2024Shift1]

Options:

Answer:D

Solution:



-------------------------------------------------------------------------------------------------

Question5

InanA.P.,thesixthtermsa6=2.Ifthea1a4a5isthegreatest,thenthe

commondifferenceoftheA.P.,isequalto

[29-Jan-2024Shift1]

Options:

3/2

8/5

2/3

5/8

Answer:B

Solution:



-------------------------------------------------------------------------------------------------

Question6

Iflogea,logeb,logecareinanA.P.andlogea−loge2b,loge2b−loge3c,

loge3c−logeaarealsoinanA.P,thena:b:cisequalto

[29-Jan-2024Shift2]

Options:

9:6:4

16:4:1

25:10:4

6:3:2

Answer:A

Solution:

Question7

Ifeachtermofageometricprogressiona1,a2,a3,...witha1=1/8anda2≠

a1,isthearithmeticmeanofthenexttwotermsandSn=a1+a2+...+an,

thenS20−S18isequalto

[29-Jan-2024Shift2]

Options:

215

−218

218

−215

Answer:D

Solution:

Question8

LetSadenotethesumoffirstntermsanarithmeticprogression.IfS20=790

andS10=145,thenS15−S5is:

[30-Jan-2024Shift1]

Options:

395

390

405

410

Answer:A

Solution:

Question9

Letα=12+42+82+132+192+262+.......upto10termsand If4α

−β=55k+40,thenkisequalto

[30-Jan-2024Shift1]

Options:

Answer:353

Solution:

Question10

Letaandbbebetwodistinctpositiverealnumbers.Let11thtermofaGP,

whosefirsttermisaandthirdtermisb,isequaltopthtermofanotherGP,

whosefirsttermisaandfifthtermisb.Thenpisequalto

[30-Jan-2024Shift2]

Options:

Answer:C

Solution:



-------------------------------------------------------------------------------------------------

Question11

LetSnbethesumton-termsofanarithmeticprogression3,7,11,.......

[30-Jan-2024Shift2]

Options:

Answer:9

Solution:



-------------------------------------------------------------------------------------------------

Question12

Thesumoftheseries upto10termsis

[31-Jan-2024Shift1]

Options:

45/109

55/109

Answer:D

Solution:



-------------------------------------------------------------------------------------------------

Question13

Let2nd,8thand44th,termsofanon-constantA.P.berespectivelythe1st,

2ndand3rdtermsofG.P.IfthefirsttermofA.P.is1thenthesumoffirst20

termsisequalto-

[31-Jan-2024Shift2]

Options:

980

960

990

970

Answer:D

Solution:



-------------------------------------------------------------------------------------------------

Question14

Let3,a,b,cbeinA.P.and3,a−1,b+1,c+9beinG.P.Then,thearithmetic

meanofa,bandcis:

[1-Feb-2024Shift1]

Options:

-4

-1

Answer:D

Solution:



-------------------------------------------------------------------------------------------------

Question15

Let3,7,11,15,....,403and2,5,8,11,..,404betwoarithmeticprogressions.

Thenthesum,ofthecommontermsinthem,isequalto____

[1-Feb-2024Shift1]

Options:

Answer:6699

Solution:



-------------------------------------------------------------------------------------------------

Question16

LetSndenotethesumofthefirstntermsofanarithmeticprogression.

IfS10=390andtheratioofthetenthandthefifthtermsis15:7,thenS15−

S5isequalto:

[1-Feb-2024Shift2]

Options:

800

890

790

690

Answer:C

Solution:



-------------------------------------------------------------------------------------------------

Question17

IfthreesuccessivetermsofaG.P.withcommonratior(r>1)arethelengths

ofthesidesofatriangleand[r]denotesthegreatestintegerlessthanor

equaltor,then3[r]+[−r]isequalto:

[1-Feb-2024Shift2]

Options:

Answer:1

Solution:



Question18

Forthreepositiveintegersp,q,r,xpq q2=yqr =zp2randr =pq +1suchthat

3,3logyx,3logzy,7logxzareinA.P.withcommondifference 1

2.Thenr −p−q

isequalto

[24-Jan-2023Shift1]

Options:

A.2

B.6

C.12

D.−6

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question19

The4 th termofGPis500anditscommonratiois 1

m,m∈N.LetSndenote

thesumofthefirstntermsofthisGP.IfS6>S5+1andS7<S6+1

2,thenthe

numberofpossiblevaluesofmis

[24-Jan-2023Shift1]

Answer:12

Solution:

pq2=logxλ

qr =logyλ

p2r=logzλ

logyx=qr

pq2=r

pq... . (1)

logxz=pq2

p2r=q2

pr..........(2)

logzy=p2r

qr =p2

q..... . (3)

3,3r

pq,3p2

q,7q2

pr inA.P

pq −3=1

2

r=7

6pq......(4)

r=pq +1

pq =6......(5)

r=7..... . (6)

3p2

q=4

Aftersolvingp=2andq=3

-------------------------------------------------------------------------------------------------

Question20

If 13+23+33+ ...... upto n terms

1⋅3+2⋅5+3⋅7+ ...... upto n terms =9

5,thenthevalueofnis

[24-Jan-2023Shift2]

Answer:5

Solution:

-------------------------------------------------------------------------------------------------

Question21

LetA1,A2,A3bethethreeA.P.withthesamecommondifferencedand

havingtheirfirsttermsasA,A+1,A+2,respectively.Leta,b,cbethe

T4=500 where a =firstterm,

r=commonratio =1

m,m∈N

ar3=500

m3=500

Sn−Sn−1=arn−1

S6>S5+1 and S7−S6<1

S6−S5>1a

m6<1

ar5>1 m3>103

500

m2>1 m >10

m2<500 ......(1)

From(1)and(2)

m=11,12,13............., 22

Sonumberofpossiblevaluesofmis12

13+23+33..... + n3=n(n+1)

1⋅3+2⋅5+3⋅7+ ... + nn+terms =

∑

r=1

r(2r +1) =

∑

r=1

(2r2+r)

=2⋅n(n+1)(2n +1)

6+n(n+1)

=n(n+1)

6(2(2n +1) + 3)

=n(n+1)

2×(4n +5)

=n2(n+1)2

⇒

5n(n+1)

2×(4n +5)

2=9

5

⇒5n(n+1)

2=9(4n +5)

⇒15n(n+1) = 18(4n +5)

⇒15n2+15n =72n +90

⇒15n2−57n −90 =0⇒5n2−19n −30 =0

⇒ (n−5)(5n +6) = 0

⇒n=−6

5or 5

⇒n=5.

( )

7th ,9th ,17 th termsofA1,A2,A3,respectivelysuchthat

a 7 1

2b 17 1

c 17 1

+70 =0

Ifa =29,thenthesumoffirst20termsofanAPwhosefirsttermisc −a−b

andcommondifferenceis d

12,isequalto________.

[25-Jan-2023Shift1]

Answer:495

Solution:

-------------------------------------------------------------------------------------------------

Question22

Forthetwopositivenumbersa,b,ifa,band 1

18areinageometric

progression,while 1

a,10and 1

bareinanarithmeticprogression,then,

16a +12bisequalto________.

[25-Jan-2023Shift2]

Answer:3

Solution:

| |

A+6d 7 1

2(A+1+8d)17 1

A+2+16d 17 1

+70 =0

⇒A= −7 and d =6

∴c−a−b=20

S20 =495

| |

a,b,1

18 →GP

18 =b2

a,10,1

b→AP

a+1

b=20

⇒a+b=20 ab,fromeq.(i) ;weget

⇒18b2+b=360b3

⇒360b2−18b −1=0{∵b≠0}

⇒b=18 ± √324 +1440

720

⇒b=18 + √1764

720 {∵b>0}

⇒b=1

⇒=18×1

144 =1

8

-------------------------------------------------------------------------------------------------

Question23

Leta1,a2,a3, ....beaGPofincreasingpositivenumbers.Iftheproductof

fourthandsixthtermsis9andthesumoffifthandseventhtermsis24,then

a1a9+a2a4a9+a5+a7isequalto______.

[29-Jan-2023Shift1]

Answer:60

Solution:

-------------------------------------------------------------------------------------------------

Question24

Let{ak}and{bk}, k∈N,betwoG.P.swithcommonratior1andr2

respectivelysuchthata1=b1=4andr1<r2.Letck=ak+bk,k∈N.Ifc2=5

andc3=13

4then ∞

∑

k=1ck− (12a6+8b4)isequalto__

[29-Jan-2023Shift2]

Answer:9

Solution:

-------------------------------------------------------------------------------------------------

Question25

Now,16a +12b =16×1

8+12×1

12 =3

a4⋅a6=9⇒ (a5)2=9⇒a5=3

&a5+a7=24 ⇒a5+a5r2=24 ⇒ (1+r2) = 8⇒r= √7

⇒a=3

⇒a1a9+a2a4a9+a5+a7=9+27 +3+21 =60

Giventhat

ck=ak+bkand a1=b1=4

also a2=4r1a3=4r1

b2=4r2b3=4r2

Nowc2=a2+b2=5andc3=a3+b3=13

⇒r1+r2=5

4andr1

2+r2

2=13

Hencer1r2=3

8whichgivesr1=1

2&r2=3

∑

k=1∞c

− (12a6+8b4)

1−r1

1−r2

−48

32 +27

=24 −15 =9

( )

Leta1=b1=1andan=an−1+ (n−1), bn=bn−1+an−1, ∀n≥2.IfS =∑

n=110 bn

andT =∑

n=18n

n−1

,then27(2S −T)isequalto______.

[29-Jan-2023Shift2]

Answer:461

Solution:

-------------------------------------------------------------------------------------------------

Question26

Ifan=−2

4n2−16n +15,thena1+a2+ ...... + a25isequalto:

[30-Jan-2023Shift1]

Options:

A. 51

144

B. 49

138

C. 50

141

D. 52

147

As, S =b1

2+b2

22+ ...... + b9

29+b10

210

⇒S

2=b1

22+b2

23+ ...... + b9

210 +b10

211

subtracting

⇒S

2=b1

2+a1

22+a2

23...... + a9

210 −b10

211

⇒S=b1−b10

210 +a1

2+a2

22........ + a9

⇒S

2=b1

2−b10

211 +a1

22+a2

23....... + a9

210

subtracting 

⇒s

2=b1

2−b10

211 +a1

2−a9

210 +1

22+2

23+ ... + 8

⇒S

2=a1+b1

2−(b10 +2a9)

211 +T

⇒2S =2(a1+b1) − (b10 +2a9)

29+T

⇒27(2S −T) = 28(a1+b1) − (b10 +2a9)

4

Also, bn−bn−1=an−1

∴b10 −b1=a1+a2+ ... + a9

=1+2+4+7+11 +16 +22 +29 +37

⇒b10 =130(As b1=1)

∴27(2S −T) = 28(1+1) − (130 +2×37)

29−204

4=461

( )

( ) ( )

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question27

Let ∞

∑

n=0

n3((2n)!) + (2n −1)(n!)

(n!)((2n)!) =ae +b

e+c,wherea,b,c∈ ℤande =∞

∑

n=0

nThen

a2−b+cisequalto_______.

[30-Jan-2023Shift1]

Answer:26

Solution:

-------------------------------------------------------------------------------------------------

Question28

Leta,b,c>1,a3,b3andc3beinA.P.,andlogab,logcaandlogbcbeinG.P.If

thesumoffirst20termsofanA.P.,whosefirsttermis a+4b +c

3andthe

commondifferenceis a−8b +c

10 is−444,thenabcisequalto

[30-Jan-2023Shift2]

Options:

A.343

Option(3)

Ifan=−2

4n2−16n +15thena1+a2+ ........ . a25

⇒

∑

n=1

an= ∑ −2

4n2−16n +15

= ∑ −2

4n2−6n −10n +15

= ∑ −2

2n(2n −3) − 5(2n −3)

= ∑ −2

(2n −3)(2n −5)

= ∑ 1

2n −3−1

2n −5

47 −1

(−3)

=50

141

∞

∑

n=0

n3((2n)!) + (2n −1)(n!)

(n!)((2n)!)

∞

∑

n=0

(n−3)! +

∞

∑

n=0

(n−2)!

∞

∑

n=0

(n−1)! +

∞

∑

n=0

(2n −1)! −

∞

∑

n=0

(2n)!

=e+3e +e+1

2e−1

e−1

2e+1

=5e −1

∴a2−b+c=26

( ) ( )

B.216

C. 343

D. 125

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question29

Theparabolas:ax2+2bx +cy =0andd x2+2ex +f y =0intersectontheline

y=1.Ifa,b,c,d,e,f arepositiverealnumbersanda,b,careinG.P.,then

[30-Jan-2023Shift2]

Options:

A.d,e,fareinA.P.

B. d

a,e

b,f

careinG.P.

C. d

a,e

b,f

careinA.P.

D.d,e,fareinG.P.

Answer:C

Solution:

Asa3,b3,c3beinA.P.→a3+c3=2b3...(1)

loga

b,logc

a,logb

careinG.P.

∴log b

log a ⋅log c

log b =log a

log c

∴ (log a)3= (log c)3⇒a=c...(2)

From(1)and(2)

a=b=c

T1=a+4b +c

3=2a;d=a−8b +c

10 =−6a

10 =−3

∴S20 =20

24a +19 −3

=10 20a −57a

= −74a

∴ −74a = −444 ⇒a=6

∴abc =63=216

( )

[ ( ) ]

[ ]

ax2+2bx +c=0

⇒ax2+2√acx +c=0(∵b2=ac)

⇒ (x√a+ √c)2=0

x2−√c

√a......

Now, dx2+2ex +f=0

⇒dc

a+2e −√c

√a+f=0

⇒d c

a+f=2e c

a

⇒d

a+f

c=2e 1

⇒d

a+f

c=2e

b[as b = √ae]

( ) [ ]

√

-------------------------------------------------------------------------------------------------

Question30

The8 th commontermoftheseries

S1=3+7+11 +15 +19 + ....

S2=1+6+11 +16 +21 + ...

is_______.

[30-Jan-2023Shift2]

Answer:151

Solution:

-------------------------------------------------------------------------------------------------

Question31

Lety =f(x)representaparabolawithfocus −1

2,0 anddirectrixy = − 1

Then

S=x∈R:tan−1(√f(x)+sin −1(√f(x) + 1)) = π

2:

[31-Jan-2023Shift1]

Options:

A.containsexactlytwoelements

B.containsexactlyoneelement

C.isaninfiniteset

D.isanemptyset

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question32

( )

{ }

∴d

a,e

b,f

careinA.P.

T8=11 + (8−1) × 20

=11 +140 =151

x+1

2=y+1

y= (x2+x)

tan−1√x(x+1)+sin −1x2+x+1=π∕2

0≤x2+x+1≤1

x2+x≤0...(1)

Also x2+x≥0...(2)

∴x2+x=0⇒x=0, −1

Scontains2element.

( ) ( )

√

Leta1,a2, ......, anbeinA.P.Ifa5=2a7anda11 =18,then

12 1

√a10 + √a11

√a11 + √a12

+ .... 1

√a17 + √a18

isequalto_______.

[31-Jan-2023Shift1]

Answer:8

Solution:

-------------------------------------------------------------------------------------------------

Question33

Leta1,a2,a3, ....beanA.P.Ifa7=3,theproducta1a4isminimumandthe

sumofitsfirstntermsiszero,thenn! − 4an(n+2)isequalto:

[31-Jan-2023Shift2]

Options:

A.24

B. 33

C. 381

D.9

Answer:A

Solution:

( )

2a7=a5(given)

2(a1+6d ) = a1+4d

a1+8d =0...(1)

a1+10d =18...(2)

By (1)and (2)weget a1= −72,d=9

a18 =a1+17d = −72 +153 =81

a10 =a1+9d =9

12 √a11 − √a10

d+√a12 − √a11

d+ ...... √a18 − √a17

12 √a18 − √a10

d=12(9−3)

9=12 ×6

6=8

( )

a+6d =3...(1)

Z=a(a+3d )

= (3−6d )(3−3d )

=18d 2−27d +9

Differentiatingwithrespecttod

⇒36d −27 =0

⇒d=3

4,from(1)a=−3

2, ( Z= minimum)

Now,Sn=n

2−3+ (n−1)3

4=0

⇒n=5

Now,

n! − 4an(n+2)=120 −4(a35)

=120 −4(a+ (35 −1)d)

( )

-------------------------------------------------------------------------------------------------

Question34

Thesum

12−2.32+3.52−4.72+5.92− ........ + 15.292is________.

[31-Jan-2023Shift2]

Answer:6952

Solution:

-------------------------------------------------------------------------------------------------

Question35

Thesumto10termsoftheseries 1

1+12+14+2

1+22+24+3

1+32+34+ ....is:-

[1-Feb-2023Shift1]

Options:

A. 59

111

B. 55

111

C. 56

111

D. 58

111

Answer:B

Solution:

=120 −4−3

2+34 ⋅3

=120 −4−6+102

=120 −96 =24

( ( ) )

( )

Separatingoddplacedandevenplacedtermsweget

S= (1.12+3.52+ .. . .15 ⋅ (29)2) − ( 2.32+4.72.

+ ... + 14 . (27)2

∑

n=1

(2n −1)(4n −3)2−

∑

n=1

(2n)(4n −1)2

Applyingsummationformulaweget

=29856 −22904 =6952

Tr=(r2+r+1) − (r2−r+1)

2(r4+r2+1)

⇒Tr=1

r2−r+1−1

r2+r+1

T1=1

1−1

T2=1

3−1

T3=1

7−1

[ ]

-------------------------------------------------------------------------------------------------

Question36

Leta1=8,a2,a3, ... . anbeanA.P.Ifthesumofitsfirstfourtermsis50and

thesumofitslastfourtermsis170,thentheproductofitsmiddletwo

termsis______.

[1-Feb-2023Shift1]

Answer:754

Solution:

-------------------------------------------------------------------------------------------------

Question37

Thenumberof3-digitnumbers,thataredivisiblebyeither2or3butnot

divisibleby7is_______.

[1-Feb-2023Shift1]

Answer:514

Solution:

-------------------------------------------------------------------------------------------------

Question38

Whichofthefollowingstatementsisatautology?

[1-Feb-2023Shift2]

Options:

T10 =1

91 −1

111

⇒

∑

r=1

Tr=1

21−1

111 =55

111

[ ]

a1+a2+a3+a4=50

⇒32 +6d =50

⇒d=3

and, an−3+an−2+an−1+an=170

⇒32 + (4n −10) ⋅ 3=170

⇒n=14

a7=26,a8=29

⇒a7⋅a8=754

Divisibleby2→450

Divisibleby3→300

Divisibleby7→128

Divisibleby2&7→64

Divisibleby3&7→43

Divisibleby2&3→150

Divisibleby2,3&7→21

∴Totalnumbers = 450 +300 −150 −64 −43 +21 =514

A.p → (pΛ(p→q))

B.(pΛq) → (∼(p) → q) )

C.(pΛ(p→q))→ ∼ q

D.pV(pΛq)

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question39

Thesumofthecommontermsofthefollowingthreearithmeticprogressions.

3,7,11,15, ............, 399

2,5,8,11, ............., 359and

2,7,12,17, ......, 197,isequalto________.

[1-Feb-2023Shift2]

Answer:321

Solution:

-------------------------------------------------------------------------------------------------

Question40

Thesumofthefirst20termsoftheseries5 +11 +19 +29 +41 + ......is:

[6-Apr-2023shift1]

Options:

A.3450

B.3420

(i)p→ (pΛ(p→q))

(∼p)V(pΛ(∼pVq))

(∼p)V(fV(pΛq))

∼pV(pΛq) = (∼pVp)Λ(∼pVq)

= ∼ pVq

(ii)(pΛq) → (∼p→q)

∼(pΛq)V(p∨q) = t

{a,b,d}V{a,b,c} = V

Tautology

(iii)(pΛ(p→q))→ ∼ q

∼(pΛ(∼pVq))V∼q= ∼ (pΛq)V∼q= ∼ pV ∼q

Nottantology

(iv)pV (pΛq) = p

Nottautology.

3,7,11,15, ............, 399 d1=4

2,5,8,11, ............., 359 d2=3

2,7,12,17, ......, 197 d3=5

LCM(d1,d2,d3) = 60

Commontermsare47,107, 167

Sum =321

C.3520

D.3250

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question41

Leta1,a2,a3, ..., anbenpositiveconsecutivetermsofanarithmetic

progression.Ifd >0isitscommondifference,then:

lim

n→∞

√a1+ √a2

√a2+ √a3

+ ....... + 1

√an−1+ √an

is

[6-Apr-2023shift1]

Options:

A. 1

√d

B.1

C.√d

D.0

Answer:B

Solution:

√( )

Tn=5+(n−1)

2[2⋅6+ (n−2) ⋅ 2]

Tn=5+ (n−1)(n+4) = 5+n2+3n −4=n2+3n +1

Now S20 =

∑

n=1

Tn=

∑

n=1

n2+3n +1

S20 =20.21.41

6+3.20.21

2+20

S20 =2870 +630 +20

S20 =3520

n→∞

√a1− √a2

a1−a2

+√a2− √a3

a2−a3

+ ...... + √an−1− √an

an−1−an

=Lt

n→∞

√a1− √a2+ √a2+ √a3+ ...... + √an−1− √an

−d

=Lt

n→∞

√an− √a1

=Lt

n→∞

√n

√a1+ (n−1)d− √a1

√d

=Lt

n→∞

√d

n+d−d

n−√a1

√( )

( )

(√)

-------------------------------------------------------------------------------------------------

Question42

Ifgcd(m,n) = 1and

12−22+32−42+ ...... + (2021)2− (2022)2+ (2023)2=1012m2n

thenm2−n2isequalto:

[6-Apr-2023shift2]

Options:

A.180

B.220

C.200

D.240

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question43

If(20)19 +2(21)(20)18 +3(21)2(20)17 + ... + 20(21)19 =k(20)19,thenkisequal

to________:

[6-Apr-2023shift2]

Answer:400

Solution:

(1−2)(1+2) + (3−4)(3+4) + ...... + (2021 −2022)(2021 +2022) + (2023)2= (1012)m2n

⇒ (−1)[1+2+3+4+ ... + 2022] + (2023)2= (1012)m2n

⇒ (−1)(2022)(2023)

2+ (2023)2= (1012)m2n

⇒ (2023)[2023 −1011] = (1012)m2n

⇒ (2023)(1012) = (1012)m2n

⇒m2n=2023

⇒m2n= (17)2×7

m=17,n=7

m2−n2= (17)2−72=289 −49 =240

Ans.Option4

S= (20)19 +2(21)(20)18 + ...... + 20(21)19

20S=21(20)18 +2(21)9(20)17 + ....... + (21)20

Subtract

1−21

20 S= (20)19 + (21)(20)18 + (21)2(20)17 + ...... + (21)19 − (21)20

−1

20 S= (20)19 1−21

1−21

− (21)20

−1

20 S= (21)20 − (20)20 − (21)20

S= (20)21 =K(20)19 (given)

K= (20)2

=400

( )

( ) [( ) ]

( )

Question44

LetSK

1+2+ ... + K

Kand n

∑

j=1Sj

2=n

A(Bn2+Cn +D),whereA,B,C,D∈NandAhas

leastvalue.Then

[8-Apr-2023shift1]

Options:

A.A +BisdivisiblebyD

B.A +B=5(D−C)

C.A +C+DisnotdivisiblebyB

D.A +B+Disdivisibleby5

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question45

Letanbethen th termoftheseries5 +8+14 +23 +35 +50 + ....and

Sn =n

∑

k=1aK.ThenS30 −a40isequalto

[8-Apr-2023shift2]

Options:

A.11260

B.11280

C.11290

D.11310

Answer:C

Solution:

Sk=k+1

2=k2+1+2k

∴

∑

j−1

2=1

n(n+1)(2n +1)

6+n+n(n+1)

(n+1)(2n +1)

6+1+n+1

2n2+3n +1

6+n+2

2n2+9n +13

6=n

24[2n2+9n +13]

A=24,B=2,C=9,D=13

[ ]

Sn=5+8+14 +23 +35 +50 + ... + an

Sn=5+8+14 +23 +35 + ... + an

────────────────────────────

O=5+3+6+9+12 +15 + ... − an

-------------------------------------------------------------------------------------------------

Question46

Let0 <z<y<xbethreerealnumberssuchthat 1

x,1

y,1

zareinanarithmetic

progressionandx, √2y,zareinageometricprogression.If

xy +yz +zx =3

√2xyz,then3(x+y+z)2isequalto_______.

[8-Apr-2023shift2]

Answer:150

Solution:

-------------------------------------------------------------------------------------------------

Question47

Letthefirsttermaandthecommonratiorofageometricprogressionbe

positiveintegers.Ifthesumofsquaresofitsfirstthreeis33033,thenthe

sumofthesetermsisequalto:

[10-Apr-2023shift1]

Options:

A.210

B.220

an=5+ (3+6+9+ .. . (n−1)terms )

an=3n2−3n +10

a40 =3(40)2−3(40) + 10

2=2345

S30 =

n=1

n2−3

n=1

n+10

n=1

3×30 ×31 ×61

6−3×30 ×31

2+10 ×30

S30 =13635

S30 −a40 =13635 −2345

=11290(Option (3))

y=1

x+1

2y2=xz

y=x+z

xz =x+z

2y2

x+z=4y

xy +yz +zx =3

√2xyz

y(x+z) + zx =3

√2xz ⋅y

4y2+2y2=3

√2y⋅2y2

6y2=3√2y3

y= √2

x+y+z=5y =5√2

3(x+y+z)2=3×50 =150

C.231

D.241

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question48

Iff (x) = (tan 1∘)x+loge(123)

xloge(1234) − (tan 1∘),x>0,thentheleastvalueoff (f(x)) + f f 4

xis:

[10-Apr-2023shift1]

Options:

A.2

B.4

C.8

D.0

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question49

( ( ) )

Leta,ar,ar2bethreetermsofGP

Given:a2+ (ar)2+ (ar2)2=33033

a2(1+r2+r4) = 112.3.7.13

⇒a=11 and 1 +r2+r4=3.7.13

⇒r2(1+r2) = 273 −1

⇒r2(r2+1) = 272 =16 ×17

⇒r2=16

∴r=4[∵r>0]

Sumofthreeterms = a+ar +ar2=a(1+r+r2)

=11(1+4+16)

=11 ×21 =231

f(x) = (tan 1)x+log 123

x log 1234 −tan 1

LetA=tan 1,B=log 123,C=log 1234

f(x) = Ax +B

xC −A

f(f(x)) =

AAx +B

xC −A+B

CAx +B

CX −A−A

=A2x+AB +xBC −AB

ACx +BC −ACx +A2

=x(A2+BC)

(A2+BC.=x

f(f(x)) = x

f f 4

x=4

f(f(x)) + f f 4

AM ≥GM 

x+4

x≥4

( )

( ( ) )

Thesumofallthoseterms,ofthearithmeticprogression3,8,13, ...., 373,

whicharenotdivisibleby3,isequalto_______.

[10-Apr-2023shift1]

Answer:9525

Solution:

-------------------------------------------------------------------------------------------------

Question50

IfSn=4+11 +21 +34 +50 + ....tonterms,then 1

60(S29 −S9)isequalto

[10-Apr-2023shift2]

Options:

A.220

B.227

C.226

D.223

Answer:D

Solution:

A.P:3,8,13... . .373

Tn=a+ (n−1)d

373 =3+ (n−1)5

⇒n=370

⇒n=75

NowSum =n

2[a+1]

=75

2[3+373] = 14100

Nownumbersdivisibleby3are,

3,18,33..... . .363

363 =3+ (k−1)15

⇒k−1=360

15 =24 ⇒k=25

Now,sum =25

2(3+363) = 4575s

∴req.sum =14100 −4575

=9525

Sn=4+11 +21 +34 +50 + ... + nterms

DifferenceareinA.P.

LetTn=an2+bn +c

T1=a+b+c=4

T2=4a +2b +c=11

T3=9a +3b +c=21

Bysolvingthese3equations

a=3

2,b=5

2,c=0

SoTn=3

2n2+5

Sn=ΣT n

2∑n2+5

2∑n

-------------------------------------------------------------------------------------------------

Question51

Supposea1,a2,2,a3,a4beinanarithemetico-geometricprogression.Ifthe

commonratioofthecorrespondinggeometricprogressionin2andthesum

ofall5termsofthearithmetico-geometricprogressionis 49

2,thena4isequal

to_______.

[10-Apr-2023shift2]

Answer:16

Solution:

-------------------------------------------------------------------------------------------------

Question52

Letx1,x2, ....., x100beinanarithmeticprogression,withx1=2andtheir

meanequalto200.Ifyi=i(xi−i),1 ≤i≤100,thenthemeanof

y1,y2, .... . y100is:

[11-Apr-2023shift1]

Options:

A.10051.50

B.10100

C.10101.50

D.10049.50

Answer:D

Solution:

n(n+1)(2n +1)

6+5

(n)(n+1)

=n(n+1)

4[2n +1+5]

Sn=n(n+1)

4(2n +6) = n(n+1)(n+3)

29 ×30 ×32

2−9×10 ×12

2=223

( )

(a−2d )

4,(a−d)

2,a,2(a+d), 4(a+2d )

a=2

4+1

2+1+6×2+ (−1+2+8)d=49

4+7+9d =49

9d =49

2−62

4=98 −62

4=9

d=1

⇒a4=4(a+2d )

=16

( )

-------------------------------------------------------------------------------------------------

Question53

LetS =S=109 +108

5+107

52+ .... + 2

5107 +1

5108.Thenthevalueof(16S − (25)−54)

isequalto_______.

[11-Apr-2023shift1]

Answer:2175

Solution:

-------------------------------------------------------------------------------------------------

Question54

Leta,b,canddbepositiverealnumberssuchthata +b+c+d=11.Ifthe

maximumvalueofa5b3c2dis3750β,thenthevalueofβis

[11-Apr-2023shift2]

Options:

A.55

Mean =200

⇒

100

2(2×2+99d)

100 =200

⇒4+99d =400

⇒d=4

yi=i(xi −1)

=i(2+ (i−1)4−i) = 3i2−2i

Mean =Σyi

100

∑

i=1

3i2−2i

100

3×100 ×101 ×201

6−2×100 ×101

=101 201

2−1=101 ×99.5

=10049.50

{ }

S=109 +108

5+107

52...... + 1

5108

5=109

5+108

52...... ⋅ 2

5108 +1

5109

5=109 −1

5−1

52...... − 1

5108 −1

5109

=109 −1

1−1

5109

1−1

=109 −1

41−1

5109

=109 −1

4+1

4×1

5109

S=5

4109 −1

4+1

4.5109

16S =20 ×109 −5+1

5108

16S − (25)−54 =2180 −5=2175

(( ) )

( ) )

( )

B.108

C.90

D.110

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question55

Fork ∈N,ifthesumoftheseries1 +4

k+8

k2+13

k3+19

k4+ ......is10,thenthe

valueofkis______

[11-Apr-2023shift2]

Answer:2

Solution:

Givena+b+c+d=11

(a,b,c,d>0}

(a5b3c2d)max.=?

LetassumeNumbers-

5,a

5,b

3,b

3,c

2,c

WeknowA.M.≥G.M.

5+a

5+b

3+b

3+c

2+c

2+d

11 ≥a5b3c2d

55⋅33⋅22⋅1

11 ≥a5b3c2d

55⋅33⋅22⋅1

a5⋅b3⋅c2⋅d≤55⋅33⋅22

max(a5b3c2d) = 55⋅33⋅22=337500

=90 ×3750 =β×3750

β=90

Option(C)90correct

( )

10 =1+4

k+8

k2+13

k3+19

k4+ ...... .upto ∞

9=4

k+8

k2+13

k3+19

k4+ ....... upto ∞

k=4

k2+8

k3+13

k4+ ..... .upto ∞

_________________________________

S=9 1 −1

k=4

k+4

k2+5

k3+6

k4...... upto ∞

k=4

k2+4

k3+5

k4+ ..... .upto ∞

__________________________________

1−1

kS=4

k+1

k3+1

k4+1

k5+ .... . ∞

9 1 −1

2=4

1−1

9(k−1)3=4k(k−1) + 1

k=2

( )

( ) ( )

-------------------------------------------------------------------------------------------------

Question56

Let<an>beasequencesuchthata1+a2+ .. + an=n2+3n

(n+1)(n+2).If

28 10

∑

k=1

=p1p2p3...pm,wherep1,p2......pmarethefirstmprimenumbers,

thenmisequalto

[12-Apr-2023shift1]

Options:

A.8

B.5

C.6

D.7

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question57

Lets1,s2,s3, ......., s10respectivelybethesumto12termsof10A.P.sm

whosefirsttermsare1,2,3, ..., 10andthecommondifferencesare

1,3,5, .........., 19respectively.Then 10

∑

i=1siisequalto

[13-Apr-2023shift1]

Options:

A.7260

B.7380

C.7220

D.7360

Answer:A

Solution:

an=Sn−Sn−1=n2+3n

(n+1)(1+2)−(n−1)(n+2)

n(n+1)

⇒an=4

n(n+1)(1+2)

⇒28

∑

k−1

=28

∑

k−1

k(k+1)(k+2)

∑

k−1

(k(k+1)(k+2)(k+3) − (k−1)k(k+1)(k+2))

4⋅10 ⋅11 ⋅12 ⋅13 =2⋅3⋅5⋅7⋅11 ⋅13

Som=6

Sk=6(2k + (11)(2k −1))

Sk=6(2k +22k −11)

Sk=144k −66

-------------------------------------------------------------------------------------------------

Question58

Thesumto20termsoftheseries2.22−32+2.42−52+2.62− ........isequal

to_______.

[13-Apr-2023shift1]

Answer:1310

Solution:

-------------------------------------------------------------------------------------------------

Question59

Leta1,a2,a3, ....beaG.P.ofincreasingpositivenumbers.Letthesumofits

6th and8 th termsbe2andtheproductofits3rdand5 th termsbe 1

9.Then

6(a2+a4)(a4+a6)isequalto

[13-Apr-2023shift2]

Options:

A.2

B.3

C.3√3

D.2√2

Answer:B

Solution:

∑

Sk=144

∑

k=1

k−66 ×10

=144×10 ×11

2−660

=7920 −660

=7260

(22−32+42−52+20 terms ) + (22+42+ .... + 10 terms )

− (2+3+4+5+ .... + 11) + 4[1+22+ ... . .102]

−21 ×22

2−1+4×10 ×11 ×21

=1−231 +14 ×11 ×10

=1540 +1−231

=1310

[ ]

a3⋅a5=1

⇒ar2⋅ar4=1

⇒ (ar3)2=1

⇒ar3=1

3...(i)

a6+a8=2

⇒ar5+ar7=2

-------------------------------------------------------------------------------------------------

Question60

Let[α]denotethegreatestinteger≤α.Then[√1] + [√2] + [√3] + ... + [√120]

isequalto_______.

[13-Apr-2023shift2]

Answer:825

Solution:

-------------------------------------------------------------------------------------------------

Question61

Letf (x) = 10

∑

k=1kxk,x∈RIf2f ′(2) − f′(2) = 119(2)n+1thennisequalto_______

[13-Apr-2023shift2]

Answer:10

Solution:

⇒ar3(r2+r4) = 2

⇒1

3r2(1+r2) = 2

⇒r2(1+r2) = 2×3

⇒r2=2⇒r= √2

a=1

3×1

2√2=1

6√2

6(a2+a4)(a4+a6)

⇒6(ar +ar3)(ar3+ar5)

⇒6ar3

r2+1

3+1

3r2=3

( ) ( )

S= [√1] + [√2] + [√3] + ... + [√120]

[√1] → [√3] = 1×3

[√4] → [√8] = 2×5

[√9] → [√15] = 3×7

⋮

[√100] → [√120] = 10 ×21

S=1×3+2×5+3×7+ ... + 10 ×21

∑

r=1

r(2r +1)

∑

r=1

r2+

∑

r=1

=2×10 ×11 ×21

6+10 ×11

=770 +55

=825

f(x) =

∑

k=1

kx xk

⇒f(x) = x+2x2+3x3+ ⋯ +9x9+10x10 − (i)

-------------------------------------------------------------------------------------------------

Question62

LetA1andA2betwoarithmeticmeansandG1,G2,G3bethreegeometric

meansoftwodistinctpositivenumbers.ThenG1

4+G2

4+G3

4+G1

2G3

2is

equalto

[15-Apr-2023shift1]

Options:

A.2(A1+A2)G1G3

B.(A1+A2)2G1G3

C.2(A1+A2)G1

2G3

D.(A1+A2)G1

2G3

Answer:B

Solution:

xf(x) = x2+2x3+ ... + 9x10 +10x11... (ii)

"(i) − (ii)1

f(x)(1−x) = x+x2+x3+ ⋯ +x10 −10x11

f(x)(1−x) = x(1−x10)

1−x−10x11

f(x) = x(1−x10)

(1−x)2−10x11

(1−x)

f(2) = 2+g(2)11

(1−x)2f(x) = x(1−x10) − 10x11(1−x)

diff.w.r.t. x

(1−x)2f′(2) + f(2)2(1−x)(−1)

=x(−10x9) + (1−x10) − 10x11(−1) − (1−x)(110)x10

put x =2

f′(2) + f(2)(2) = −10(2)10 +1−210 +10(2)11 −110(2)10 +110(2)11

= (−121)210 + (120)211 +1

=210(240 −121) + 1

=119(2)10 +1

n=10

a,A1,A2,bareinA.P.

d=b−a

3;A1=a+b−a

3=2a +b

A2=a+2b

A1+A2=a+b

a,G1,G2,G3,bareinG.P.

r=b

G1= (a3b)

G2= (a2b2)

G3= (ab3)

4

4+2

4+G3

4+G1

2G3

a3b+a2b2+ab3+ (a3b)

2⋅ (ab3)

=a3b+a2b2+ab3+a2⋅b2

=ab(a2+2ab +b2)

=ab(a+b)2

( )

-------------------------------------------------------------------------------------------------

Question63

Ifthesumoftheseries 1

2−1

3+1

22−1

2⋅3+1

32+1

23−1

22⋅3+1

2⋅32−1

33+

24−1

22⋅3+1

22⋅32−1

2⋅32+1

34+ .......is α

β,whereαandβareco-prime,then

α+3βisequalto_______

[15-Apr-2023shift1]

Answer:7

Solution:

-------------------------------------------------------------------------------------------------

Question64

If{ai}n

i=1,wherenisaneveninteger,isanarithmeticprogressionwith

commondifference

1,and n

∑

i=1ai=192,n∕2

∑

i=1a2i =120,thennisequalto:

[24-Jun-2022-Shift-1]

Options:

A.48

B.96

C.92

D.104

Answer:B

Solution:

( ) ( ) ( )

( )

=G1⋅G3⋅ (A1+A2)2

2−1

3+1

22−1

2⋅3+1

32+1

23+1

22⋅3+1

2⋅32−1

33+ .. . P1

2+1

3=1

22−1

32+1

23+1

33+1

24−1

34+ ...

1−1

−

1+1

6=1

2−1

12 =5

∴P=1

2=α

β∴α=1,β=2

α+3β =7

( ) ( ) ( ) ( ) ( ) ( ) ( )

Question65

If 1

2⋅310 +1

22⋅39+ ... + 1

210 ⋅3=K

210 ⋅310,thentheremainderwhenKisdividedby6is

[25-Jun-2022-Shift-1]

Options:

------------------------------------------------------------------------------------------------

A.1

B.2

C.3

D.5

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question66

Thegreatestintegerlessthanorequaltothesumoffirst100termsofthe

sequence 1

3,5

9,19

27,65

81,.isequalto

[25-Jun-2022-Shift-1]

Options:

A.

Answer:98

Solution:

-------------------------------------------------------------------------------------------------

Question67

Thesum1 +2⋅3+3⋅32+ ......... + 10.39isequalto

[25-Jun-2022-Shift-2]

Options:

A. 2⋅312 +10

B. 19 ⋅310 +1

C.5 ⋅310 −2

D. 9⋅310 +1

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question68

LetA =10

∑

i=1

∑

j=1min{i,j}andB =10

∑

i=1

∑

j=1max{i,j}.ThenA +Bisequalto

[26-Jun-2022-Shift-1]

Answer:1100

Solution:

LetS=1.30+2.31+3.32+ ...... + 10.39

3S =1.31+2.32+ ......... + 10.310

−2S = (1.30+1.31+1.32+ ...... + 1.39) − 10.310

⇒S=1

210.310 −310 −1

−3−1

⇒S=19.310 +1

[ ]

-------------------------------------------------------------------------------------------------

Question69

IfA =∞

∑

n=1

(3+ (−1)n)nandB =∞

∑

n=1

(−1)n

(3+ (−1)n)n,then A

Bisequalto:

[26-Jun-2022-Shift-2]

Options:

A. 11

B.1

C.−11

D.−11

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question70

Ifa1(>0), a2,a3,a4,a5areinaG.P.,a2+a4=2a3+1and3a2+a3=2a4,then

a2+a4+2a5isequalto____

[26-Jun-2022-Shift-2]

Answer:40

Solution:

-------------------------------------------------------------------------------------------------

Question71

x=∞

∑

n=0an,y=∞

∑

n=0bn,z=∞

∑

n=0cn,wherea,b,careinA.P.and

|a| < 1, | b| <1, | c| <1,abc≠0,then:

[27-Jun-2022-Shift-1]

Options:

A.x,y,zareinA.P.

B.x,y,zareinG.P.

C. 1

x,1

y,1

zareinA.P.

D. 1

x+1

y+1

z=1− (a+b+c)

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question72

Ifthesumofthefirsttentermsoftheseries

5+2

65 +3

325 +4

1025 +5

2501 + ...

is m

n,wheremandnareco-primenumbers,thenm +nisequalto

[27-Jun-2022-Shift-1]

Answer:276

Solution:

-------------------------------------------------------------------------------------------------

Question73

LetS =2+6

7+12

72+20

73+30

74+ .....Then4Sisequalto

[27-Jun-2022-Shift-2]

Options:

A. 7

B. 73

C. 7

D. 72

Answer:C

( )

∞

∑

n=0

an=1

1−a;y=

∞

∑

n=0

bn=1

1−b;z=

∞

∑

n=0

cn=1

1−c

Now,

a,b,c→AP

1−a,1−b,1−c→AP

1−a,1

1−b,1

1−c→H P

x,y,z→HP

∴1

x,1

y,1

z→AP

Tr=r

(2r2)2+1

(2r2+1)2− (2r)2

(2r2+2r +1)(2r2−2r +1)

S10 =1

∑

r=1

(2r2−2r +1)−1

(2r2+2r +1)

41−1

5+1

5−1

13 + ... + 1

181 −1

221

⇒S10 =1

4⋅220

221 =55

221 =m

∴m+n=276

( )

[ ]

Solution:

-------------------------------------------------------------------------------------------------

Question74

Ifa1,a2,a3......andb1,b2,b3......areA.P.,and

a1=2,a10 =3,a1b1=1=a10b10,thena4b4isequalto-

[27-Jun-2022-Shift-2]

Options:

A. 35

B.1

C. 27

D. 28

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

S=2+6

7+12

72+20

73+30

74+ .........(i)

7S=2

7+6

72+12

73+20

74+ .........(ii)

(i)-(ii)

7S=2+4

7+6

72+8

73+ ............(iii)

72S=2

7+4

72+6

73+ ............(iv)

(iii)-(iv)

2S=2+2

7+2

72+2

73+ .......

=21

1−1

=27

∴4S =87

3=7

( )

[ ] ( )

( ) ( )

a1,a2,a3...areinA.P.(Letcommondifferenceisd1)

b1,b2,b3...areinA.P.(Letcommondifferenceisd2)

anda1=2,a10 =3,a1b1=1=a10b10

∵a1b1=1

∴b1=1

a10b10 =1

∴b10 =1

Now,a10 =a1+9d 1⇒d1=1

b10 =b1+9d 2⇒d2=1

3−1

2= − 1

Now,a4=2+3

9=7

b4=1

2−3

54 =4

9

∴a4b4=28

[ ]

Question75

LetA1,A2,A3, ......beanincreasinggeometricprogressionofpositivereal

numbers.IfA1A3A5A7=1

1256andA2+A4=7

36,thenthevalueofA6+A8+A10is

equalto

[28-Jun-2022-Shift-1]

Options:

A.33

B.37

C.43

D.47

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question76

Ifnarithmeticmeansareinsertedbetweenaand100suchthattheratioof

thefirstmeantothelastmeanis1 :7anda+n =33,thenthevalueofnis:

[28-Jun-2022-Shift-2]

Options:

A.21

B.22

C.23

D.24

Answer:C

Solution:

A1⋅A3⋅A5⋅A7=1

1296

(A4)4=1

1296

A4=1

A2+A4=7

A2=1

A6=1

A8=6

A10 =36

A6+A8+A10 =43

a,A1,A2.......... . An,100

LetdbethecommondifferenceofaboveA.P.then

a+d

100 −d=1

⇒7a +8d =100....... (i)

anda+n=33

-------------------------------------------------------------------------------------------------

Question77

Letforn =1,2, ......, 50,Snbethesumoftheinfinitegeometricprogression

whosefirsttermisn2andwhosecommonratiois 1

(n+1)2.Thenthevalueof

26 +50

∑

n=1Sn+2

n+1−n−1 isequalto____

[28-Jun-2022-Shift-2]

Answer:41651

Solution:

-------------------------------------------------------------------------------------------------

Question78

Let{an}n=0

∞beasequencesuchthata0=a1=0andan+2=2an+1−an+1

foralln ≥0.Then, ∞

∑

n=2

7nisequalto:

[29-Jun-2022-Shift-1]

Options:

A. 6

343

B. 7

216

C. 8

343

D. 49

216

Answer:B

Solution:

( )

and100 =a+ (n+1)d

⇒100 =a+ (34 −a)(100 −7a)

⇒800 =8a +7a2−338a +3400

⇒7a2−330a +2600 =0

⇒a=10,260

7,but a ≠260

7

∴n=23

Sn=n2

1−1

(n+1)2

=n(n+1)2

n+2= (n2+1) − 2

n+2

Now 1

26 +

∑

n=1

Sn+2

n+1−n−1

26 +

∑

n=1

(n2−n) + 21

n+1−1

n+2

26 +50 ×51 ×101

6−50 ×51

2+21

2−1

=1+25 ×17(101 −3)

=41651

( )

{ ( ) }

( )

Solution:

-------------------------------------------------------------------------------------------------

Question79

Thesumoftheinfiniteseries1 +5

6+12

62+22

63+35

64+51

65+70

66+ ......isequalto:

[29-Jun-2022-Shift-2]

Options:

A. 425

216

B. 429

216

C. 288

125

D. 280

125

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question80

Let3,6,9,12,upto78termsand5,9,13,17,upto59termsbetwoseries.

Then,thesumofthetermscommontoboththeseriesisequalto_____

[29-Jun-2022-Shift-2]

Answer:2223

an+2=2an+1−an+1&a0=a1=0

a2=2a1−a0+1=1

a3=2a2−a1+1=3

a4=2a3−a2+1=6

a5=2a4−a3+1=10

∞

∑

n=2

7n=a2

72+a3

73+a4

74+ ...

s=1

72+3

73+6

74+10

75+ ...

7s=1

73+3

74+6

75+ ...

7=1

72+2

73+3

74+ ...

49 =1

73+2

74+ ...

36s

49 =1

72+1

73+1

74+ ...

36s

49 =

1−1

36s

49 =7

49 ×6

s=7

216

Solution:

-------------------------------------------------------------------------------------------------

Question81

Leta1=b1=1,an=an−1+2andbn=an+bn−1foreverynaturalnumber

n≥sl ant2.Then 15

∑

n=1an⋅bnisequalto

[25-Jul-2022-Shift-1]

Answer:27560

Solution:

1stAP:

3,6,9,12, .......upto78terms

t78 =3+ (78 −1)3

=3+77 ×3

=234

2ndAP:

5,9,13,17, ......upto59terms

t59 =5+ (59 −1)4

=5+58 ×4

=237

Commonterm'sAP:

Firstterm = 9

CommondifferenceoffirstAP =3

AndcommondifferenceofsecondAP =4

∴Commondifferenceofcommonterms

AP =LCM(3,4) = 12

∴NewAP =9,21,33, ......

tn=9+ (n−1)12 ≤234

⇒n≤237

⇒n=19

∴S19 =19

2[2.9 + (19 −1)12]

=19(9+108)

=2223

Given,

an=an−1+2

⇒an−an−1=2

∴Inthisseriesbetweenanytwoconsecutivestermsdifferenceis2.SothisisanA.P.withcommondifference2.

Alsogivena1=1

∴Seriesis = 1,3,5,7......

∴an=1+ (n−1)2=2n −1

Alsobn=an+bn−1

Whenn=2then

b2−b1=a2=3

. ⇒ b2−1=3 [Given b1=1]

⇒b2=4

Whenn=3then

b3−b2=a3

⇒b3−4=5

⇒b3=9

∴Seriesis = 1,4,9......

=12,22,32..... . n2

∴bn=n2

Now,

∑

n=1

(an.bn)

∑

n=1

[(2n −1)n2]

-------------------------------------------------------------------------------------------------

Question82

Thesum 21

∑

n=1

(4n −1)(4n +3)isequalto

[25-Jul-2022-Shift-2]

Options:

A. 7

B. 7

C. 14

D. 21

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question83

ConsidertwoG.Ps.2,22,23, .....and4,42,43, ....of60andnterms

respectively.Ifthegeometricmeanofallthe60 +ntermsis(2)

225

8,then

∑

k=1

nk(n−k)isequalto:

[26-Jul-2022-Shift-1]

Options:

A.560

B.1540

C.1330

D.2600

Answer:C

Solution:

∑

n=1

2n3−

∑

n=1

=2(13+23+ .. . 153) − (12+22+ .. . 152)

=2×15 ×16

2−15(16) × 31

=27560

( ) ( )

∑

n=1

(4n −1)(4n +3)=3

∑

n=1

4n −1−1

4n +3

3−1

7+1

7−1

11 + ... + 1

83 −1

3−1

87 =3

3.87 =7

[ ( ) ( ) ( ) ]

[ ]

GivenG.P's2,22,23, .. . .60terms

-------------------------------------------------------------------------------------------------

Question84

Theseriesofpositivemultiplesof3isdividedintosets:

{3}, {6,9,12}, {15,18,21,24,27}, ...Thenthesumoftheelementsinthe

11 th setisequalto__________.

[26-Jul-2022-Shift-1]

Answer:6993

Solution:

-------------------------------------------------------------------------------------------------

Question85

If 10

∑

k=1

k4+k2+1=m

n,wheremandnareco-prime,thenm +nisequalto________.

[26-Jul-2022-Shift-2]

Answer:166

Solution:

4,42, ... . nterms

Now,G.M = 2

225

(2.22...4.42...)

60 +n=2

225

n2+n+1830

6+n=2

225

⇒n2+n+1830

60 +n=225

⇒8n2−217n +1140 =0

n=57

8,20,so n =20

∴

∑

k=1

k(20 −k) = 20×20 ×21

2−20 ×21 ×41

=20 ×21

220 −41

3=1330

( )

[ ]

{3×1}

1−term

, {3×2,3×3,3×4}

3 terms

, {3×5,3×6,3×7,3×8,3×9}

5−terms

, ...

∴11 th setwillhave1+ (10)2=21term

Alsoupto10 th settotal3×ktypetermswillbe1+3+5+ ...... + 19 =100−term

∴Set11 = {3×101,3×102, ..... . 3×121}

∴Sumofelements = 3× (101 +102 + ... + 121)

=3×222 ×21

2=6993

∑

k=1

k4+k2+1

∑

k=1

k2−k+1−1

k2+k+1.

[( )

-------------------------------------------------------------------------------------------------

Question86

DifferentA.P.'sareconstructedwiththefirstterm100,thelastterm199,

andintegralcommondifferences.Thesumofthecommondifferencesofall

suchA.P.'shavingatleast3termsandatmost33termsis________.

[26-Jul-2022-Shift-2]

Answer:53

Solution:

-------------------------------------------------------------------------------------------------

Question87

Supposea1,a2, ..., an, ..beanarithmeticprogressionofnaturalnumbers.If

theratioofthesumoffirstfivetermstothesumoffirstninetermsofthe

progressionis5 :17and,110 <a15 <120,thenthesumofthefirstten

termsoftheprogressionisequalto

[27-Jul-2022-Shift-1]

Options:

A.290

B.380

C.460

D.510

Answer:B

Solution:

21−1

3+1

3−1

7+1

7−1

13 + ... + 1

91 −1

111

21−1

111 =110

2.111 =55

111 =m

∴m+n=55 +111 =166

[ ]

d1=199 −100

2∉I

d2=199 −100

3=33

d3=199 −100

4∉I

dn=199 −100

i+1∈I

di=33 +11,9

SumofCD's = 33 +11 +9

=53

∵a1,a2, ... . anbeanA.Pofnaturalnumbersand

17 ⇒

2[2a1+4d ]

2[2a1+8d ]

-------------------------------------------------------------------------------------------------

Question88

Letf (x) = 2x2−x−1andS = {n∈ ℤ: | f(n) | ≤800}.Then,thevalueof ∑

n∈Sf(n)

isequalto_______.

[27-Jul-2022-Shift-1]

Answer:10620

Solution:

-------------------------------------------------------------------------------------------------

Question89

LetthesumofaninfiniteG.P.,whosefirsttermisaandthecommonratiois

r,be5.Letthesumofitsfirstfivetermsbe 98

25.Thenthesumofthefirst21

termsofanAP,whosefirsttermis10ar,nth termisanandthecommon

differenceis10ar2,isequalto:

[27-Jul-2022-Shift-2]

Options:

A.21a11

B.22a11

C.15a16

D.14a16

Answer:A

Solution:

⇒34a1+68d =18a1+72d

⇒16a1=4d

∴d=4a1

And110 <a15 <120

∴110 <a1+14d <120 ⇒110 <57a1<120

∴a1=2(∵ai∈N)

d=8

∴S10 =5[4+9×8] = 380

∵ | f(n) | ≤800

⇒−800 ≤2n2−n−1≤800

⇒2n2−n−801 ≤0

∴n∈−√6409 +1

4,√6409 +1

4andn∈z

∴n= −19, −18, −17, ........., 19,20.

∴∑ (2x2−x−1) = 2∑x2− ∑ x− ∑ 1.

=2.2 ⋅ (12+22+ ... + 192) + 2.202−20 −40

=10620

[ ]

LetfirsttermofG.P.beaandcommonratioisrThen, a

1−r=5......(i)

-------------------------------------------------------------------------------------------------

Question90

23−13

1×7+43−33+23−13

2×11 +63−53+43−33+23−13

3×15 + ⋯ + 303−293+283−273+ ... + 23−13

15 ×63 isequalto

________.

[27-Jul-2022-Shift-2]

Answer:120

Solution:

-------------------------------------------------------------------------------------------------

Question91

Considerthesequencea1,a2,a3, ...suchthata1−1,a2−2and

an+2=2

an+1

+a−nforn −1,2,3, ...If

a1+1

a3⋅

a2+1

a4⋅

a3+1

a5.. .

a30 +1

a31

a32 =2α(61C31),thenαisequal

to:

[28-Jul-2022-Shift-1]

Options:

A.−30

B.−31

C.−60

( ) ( ) ( ) ( )

a(r5−1)

(r−1)=98

25 ⇒1−r5=98

125

∴r5=27

125,r=3

∴Then, S21 =21

2[2×10ar +20 ×10ar2]

=21[10ar +10.10ar2]

=21a11

( )

Tn=

∑

k=1

[(2k)3− (2k −1)3]

n(4n +3)

∑

k=1

4k2+ (2k −1)2+2k(2k −1)

n(4n +3)

∑

k=1

(12k2−6k +1)

n(4n +3)

=2n(2n2+3n +1) − 3n2−3n +n

n(4n +3)

=n2(4n +3)

n(4n +3)=n

∴Tn=n

Sn=

∑

n=1

Tn=15 ×16

2=120

D.−61

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question92

Forp,q∈Rs considertherealvaluedfunctionf (x) = (x−p)2−q,x∈Rand

q>0.Leta1,a2′a3anda4beinanarithmeticprogressionwithmeanpand

positivecommondifference.Iff (ai) | = 500foralli =1,2,3,4,thenthe

absolutedifferencebetweentherootsoff (x) = 0is_______.

[28-Jul-2022-Shift-1]

Answer:50

Solution:

-------------------------------------------------------------------------------------------------

Question93

Letx1,x2,x3, ..., x20beingeometricprogressionwithx1=3andthecommon

an+2=2

an+1

+an

⇒anan+1+1=an+1an+2−1

⇒an+2an+1−an⋅an+1=2

For

n=1 a3a2−a1a2=2

n=2 a4a3−a3a2=2

n=3 a5a4−a4a3=2

.

n=nan+2an+1−anan+1=2

an+2an+1=2n +a1a2

Now,

(a1a2+1)

a2a3

⋅(a2a3+1)

a3a4

⋅(a3a4+1)

a4a5

......⋅ (a30a31 +1)

a31a32

4⋅5

6⋅7

8......⋅ 61

=2−60(61C31)

∵a1,a2,a3,a4

∴a2=p−3d ,a2=p−d,a3=p+danda4=p+3d

Whered>0

∵ | f(ai) | = 500

⇒ | 9d 2−q| = 500

and|d2−q| = 500

either9d 2−q=d2−q

⇒d=0notacceptable

∴9d 2−q=q−d2

∴5d 2−q=0

Rootsoff(x) = 0arep+ √qandp− √q

∴absolutedifferencebetweenroots= | 2√q| = 50

ratio 1

2.Anewdataisconstructedreplacingeachxiby(xi−i)2.Ifxisthe

meanofnewdata,thenthegreatestintegerlessthanorequaltoxis________.

[28-Jul-2022-Shift-1]

Answer:142

Solution:

-------------------------------------------------------------------------------------------------

Question94

312 +10

311 +20

310 +40

39+ ... + 10240

3=2n⋅m,wheremisodd,thenm .nisequal

to_______.

[28-Jul-2022-Shift-2]

Answer:12

Solution:

-------------------------------------------------------------------------------------------------

Question95

x1,x2,x3, ...., x20areinG.P.

x1=3,r=1

x=∑xi

2−2xii+i2

20 12 1 −1

240 −6 4 −11

218 +70 ×41

S=1+2⋅1

2+3⋅1

22+ ...

2=1

2+2

22+ ....



2=2 1 −1

220 −20

220 =4−11

218

∴[x] = 2858

20 −12

240 −66

218 ⋅1

=142

[ ( ) ( ) ]

{

( ) }

[ ( ) ]

312 +520

312 +21

311 +22

310 + ..... + 211

3=2n⋅m

⇒1

312 +51

312

((6)2−1)

(6−1)=2n⋅m

⇒1

312 +5

312 ⋅212 ⋅312 −1

312 =2n⋅m

⇒1

312 +212 −1

312 =2n⋅m

⇒2n⋅m=212

⇒m=1andn=12

m⋅n=12

( )

If 1

(20 −a)(40 −a)+1

(40 −a)(60 −a)+ ... + 1

(180 −a)(200 −a)=1

256,thenthemaximumvalueof

ais:

[29-Jul-2022-Shift-1]

Options:

A.198

B.202

C.212

D.218

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question96

Leta1,a2,a3, ...beanA.P.If ∞

∑

r=1

2r=4,then4a2isequalto________.

[29-Jul-2022-Shift-1]

Answer:16

Solution:

-------------------------------------------------------------------------------------------------

Question97

If 1

2×3×4+1

3×4×5+1

4×5×6+ ... + 1

100 ×101 ×102 =k

101,then34kisequalto________.

[29-Jul-2022-Shift-1]

20 −a−1

40 −a+1

40 −a−1

60 −a+ ... + 1

180 −a−1

200 −a=1

256

⇒1

20 −a−1

200 −a=1

256

⇒1

180

(20 −a)(200 −a)=1

256

⇒(20 −a)(200 −a) = 9.256

ORa2−220a +1696 =0

⇒a=212,8

( )

Given

S=a1

2+a2

22+a3

23+a4

24+ ..... . ∞

2S=a1

22+a2

23+ ........ . ∞

2=a1

2+(a2+a1)

22+(a3+a2)

23+ ..... . ∞

⇒S

2=a1

2+d

⇒a1+d=a2=4⇒4a2=16

Answer:286

Solution:

-------------------------------------------------------------------------------------------------

Question98

Let{an}n=0

∞beasequencesuchthata0−a1−0and

an+2=3an+1−2an+1, ∀n≥0.

Thena25a23 −2a25a22 −2a23a24 +4a22a24isequalto

[29-Jul-2022-Shift-2]

Options:

A.483

B.528

C.575

D.624

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question99

∑

r=1(r2+1)(r!)isequalto

[29-Jul-2022-Shift-2]

Options:

S=1

2×3×4+1

3×4×5+1

4×5×6+ ... + 1

100 ×101 ×102

(3−1) ⋅ 1

2×3−1

101 ×102

6−1

101 ×102

=143

102 ×101 =k

101

∴34k =286

[ ]

( )

a0=0,a1=0

an+2=3an+1−2an+1:n≥0

an+2−an+1=2(an+1−an) + 1

n=0 a2−a1=2(a1−a0) + 1

n=1 a3−a2=2(a2−a1) + 1

n=2 a4−a3=2(a3−a2) + 1

n=n an+2−an+1=2(an+1−an) + 1

(an+2−a1) − 2(an+1−a0) − (n+1) = 0

an+2=2an+1+ (n+1)

n→n−2

an−2an−1=n−1

Now a25a23 −2a25a22 −2a23a24 +4a22a24

= (a25 −2a24)(a23 −2a22) = (24)(22) = 528

A.22! − 21!

B.22! − 2(21!)

C.21! − 2(20!)

D.21! − 20!

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question100

Ife(cos2x+cos4x+cos6x+ .. . ∞)loge2satisfiestheequationt2−9t +8=0,

thenthevalueof 2 sin x

sin x + √3 cos x 0<x<π

2is

[24-Feb-2021Shift1]

Options:

A.2√3

B. 3

C.√3

D. 1

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

( )

Given,

∑

r=1

(r2+1)(r!)

Let,f(r) = (r2+1)(r!)

= (r2)(r!) + r!

=r(rr!) + r!

=r[(r+1−1)r!] + r!

=r[(r+1)r! − r!] + r!

=r[(r+1)! − (r!)] + r!

=r(r+1)! − r(r!) + r! = (r+2−2)(r+1)! − r(r!) + r!

= (r+2)(r+1)! − 2(r+1)! − [(r+1−1)(r!)] + r!

= (r+2)! − 2(r+1)! − (r+1)! + r! + r!

= (r+2)! − 3(r+1)! + 2r!

= [(r+2)! − (r+1)!] − 2[(r+1)! − r!]

∴

∑

r=1

f(r)

∑

r=1

[(r+2)! − (r+1)!] − 2

∑

r=1

[(r+1)! − r!]

= [ (22! + 21! + 20! + .... + 4! + 3!) − (21! + 20! + 19! + .... + 3! + 2!] − 2[(21! + 20! + ... + 3! + 2!) − (20! + 19! + ..... . 1!)]

= [(22!) − (2!)] − 2[(21)! − (1!)]

=22! − 2! − 2⋅ (21)! + 2.1!

=22! − 2⋅ (21)!

e(cos2x+cos4x+ .. . ∞)ℓn2 =2cos2x+cos4x+ .. . ∞=2cot2x

Nowt2−9t +9=0⇒t=1,8

⇒2cot2x=1,8⇒cot2x=0,3

⇒2 sin x

sin x + √3 cos x =2

1+ √3 cot x =2

4=1

Question101

LetA = { x:xis3-digitnumber]B = {x:x=9k +2,k∈I}and

C= {x:x=9k + ℓ, k∈I, ℓ ∈ I,O<ℓ<9}forsomeℓ(0<ℓ<9)Ifthesumof

alltheelementsofthesetA ∩ (B∪C)is274 ×400,thenℓisequalto

[24-Feb-2021Shift1]

Answer:5

Solution:

-------------------------------------------------------------------------------------------------

Question102

Thesumoffirstfourtermsofageometricprogression(G.P.)is 65

12andthe

sumoftheirrespectivereciprocalsis 65

18.Iftheproductoffirstthreetermsof

theG.P.is1andthethirdtermisα1then2αis

[2021,24Feb.Shift-II]

Answer:9

Solution:

BandCwillcontainthreedigitnumbersoftheform9k +2and9k + ℓrespectively.Weneedtofindsumofallelementsintheset

B∪Ceffectively.

Now,S(B∪C) = S(B) + S(C) − S(B∩C)whereS(k)denotessumofelementsofsetk.

Also B= {101,110, ..... . 992}

∴S(B) = 100

2(101 +992) = 54650

Case-I:Ifℓ = 2

then B ∩C=B

∴S(B∪C) = S(B)

whichisnotpossibleasgivensumis

274×400=109600

Case-II:Ifℓ ≠ 2

then B ∩C=φ

∴S(B∪C) = S(B) + S(C) = 400 ×274

⇒54650 +

110

∑

k=11

9k + ℓ = 109600

⇒9

110

∑

k=11

110

∑

k=11

ℓ = 54950

⇒9100

2(11 +110) + ℓ(100) = 54950

⇒54450 +100ℓ = 54950

⇒ℓ = 5

( )

LetfournumbersinGPbea,ar,ar2,ar3.

Accordingtothequestion,

a+ar +ar2+ar3=65

12 ⋅⋅⋅⋅⋅⋅ ⋅ (i)

and 1

a+1

ar +1

ar2+1

ar3=65

-------------------------------------------------------------------------------------------------

Question103

Thesumoftheseries ∞

∑

n=1

n2+6n +10

(2n +1)! isequalto

[2021,26Feb.Shift-II]

Options:

A. 41

8e+19

8e−1−10

B. 41

8e−19

8e−1−10

C. 41

8e+19

8e−1+10

D.−41

8e+19

8e−1−10

Answer:B

Solution:

⇒1

1+r+r2+r3

r3=65

18 ⋅⋅⋅⋅⋅⋅ ⋅ (ii)

DividingEq.(i)by(ii),weget

a(1+r+r2+r3)

(1+r+r2+r3)

=65 ∕12

65 ∕18

⇒a2r3=18

⇒a2r3=3

Also,productoffirstthreeterms = 1

a×ar ×ar2=1

⇒a3r3=1

⇒a3×3

2a2=1∵r3=3∕2

⇒a=2

3

andr3=3∕2

(2∕3)2=3

3⇒ r=3

2

Accordingtothequestion,thirdterm

α=ar2=2

3×3

2×3

2=3

2∴ 2α =2×3

2=3thirdterm

( )

[ ]

( )

Let

∞

∑

n=1

n2+6n +10

(2n +1)! =S

∞

∑

n=1

4n2+24n +40

4(2n +1)!

∞

∑

n=1

(2n +1)2+ (2n +1) ⋅ 10 +29

4(2n +1)!

∞

∑

n=1

(2n +1)2

(2n +1)(2n)! +

∞

∑

n=1

(2n +1) ⋅ 10

(2n +1)(2n)!+

∞

∑

n=1

(2n +1)!

∞

∑

n=1

(2n +1)

(2n)! +

∞

∑

n=1

(2n)! +

∞

∑

n=1

(2n +1)!  ⋅⋅⋅⋅⋅⋅ ⋅ (1)

Now,

∞

∑

n=1

(2n +1)

(2n)! =

∞

∑

n=1

(2n)! +

∞

∑

n=1

(2n)!

∞

∑

n=1

(2n −1)! +

∞

∑

n=1

(2n)!

Now,

[ ]

-------------------------------------------------------------------------------------------------

Question104

Thesumoftheinfiniteseries1 +2

3+7

32+12

33+17

34+22

35+ ...isequalto

[2021,26Feb.Shift-1]

Options:

A. 13

B. 9

C. 15

D. 11

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

∞

∑

n=1

(2n −1)! =1

1!+1

3!+1

5!+ ... =

e−1

2⋅⋅⋅⋅⋅⋅ ⋅ (ii)

and

∞

∑

n=1

(2n)! =1

2!+1

4!+1

6!+ ... =

e+1

e−2

2⋅⋅⋅⋅⋅⋅ ⋅ (iii)

and

∞

∑

n=1

(2n +1)! =1

3!+1

5!+1

7!+ ... =

e−1

e−2

2⋅⋅⋅⋅⋅⋅ ⋅ (iv)

UsingEqs.(ii),(iii),(iv)in(i),

S=1

e−1

2+11&

e+1

e−2

+29&

e−1

e−2

2

2−1

2e +11e

2+11

2e +29e

2−29

2e −4

=41e

8−19

8e −10

[ ( )

( ) ]

[ ]

Given,S=1+2

3+7

32+12

33+ ...

Let,S1=2

3+7

32+12

33+ ...

Multiply1∕3inseriesEq.(i),

3=2

32+7

33+12

34+ ...

SubtractEq.(ii)fromEq.(i),weget

S1−S1

3=2

3+5

32+5

33+ ...

⇒2S1

3=2

3+5

32+5

33+ ... 

3+5∕32

1−1∕3∵5

32+5

33+ ...isageometricserieswithr=1∕3,sumuptoinfinityofthisseriesis a

1−r,wherea= first

term]

3+5

6=9

6=3

⇒S1=3

2×3

2=9

∴S=1+S1

=1+9

4=13

[ ]

[ ] [

[ ]

Question105

Ifthearithmeticmeanandgeometricmeanofthepthandqthtermsofthe

sequence−16,8, −4,2,...satisfytheequation

4x2−9x +5=0,thenp +qisequalto

[2021,26Feb.Shift-II]

Answer:10

Solution:

-------------------------------------------------------------------------------------------------

Question106

Inanincreasinggeometricseries,thesumofthesecondandthesixthterm

is 25

2andtheproductofthethirdandfifthtermis25.Then,thesumof4th,

6thand8thtermsisequalto

[2021,26Feb.Shift-1]

Options:

A.30

B.26

IfAM andGM satisfytheequation4x2−9x +5=0,thenAMandGMarenothingbutrootsofthisquadraticequation,

4x2−9x +5=0

⇒4x2−4x −5x +5=0

⇒4x(x−1) − 5(x−1) = 0

⇒ (x−1)(4x −5) = 0

⇒x=1,5

Then,AM =5

4and GM =1[∵AM ≥GM ]

Again,thegivenseriesis

−16,8, −4,2.....

whichisageometricprogressionserieswithcommonratio −1

2,then

pthterm = −16 −1

p−1=tp

qthterm = −16 −1

q−1=tq

Arithmeticmean = 5

⇒tp+tq

2=5

4

Geometricmean = 1

⇒ √tptq=1

∵ √tptq=1

⇒(−16)−1

p−1(−16)−1

q−1=1

⇒ (−16)2−1

p+q−2=1

⇒ (−24)2−1

p+q−2=1

⇒ (−2)8(+1)p+q−2

(−2)p+q−2=1

⇒(−2)8(+1)p+q−2= (−2)p+q−2

⇒ (−2)8= (−2)p+q−2

⇒p+q−2=8

⇒p+q=10

( )

( ) ( )

( )

C.35

D.32

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question107

Theminimumvalueoff (x) = aax+a1−ax,wherea,x∈Randa >0,isequalto

[2021,25Feb.Shift-II]

Options:

A.a +1

B.a +1

C.2√a

D.2a

Answer:C

Solution:

Letthefirsttermofgeometricseriesbe'a′andcommonratiobe'r'.

Then,nthtermofgivenseriesisgivenas

Tn=arn−1

Now,giventhatsumofsecondandsixthtermis25/2.

i.e.T2+T6=25 ∕2

⇒ar +ar5=25 ∕2

⇒ar(1+r4) = 25 ∕2⋅⋅⋅⋅⋅⋅ ⋅ (i)

Also,giventhatproductofthirdandfifthtermis25.

l rl i.e. (T3)(T5) = 25

⇒ (ar2)(ar4) = 25

⇒a2r6=25 ⋅⋅⋅⋅⋅⋅ ⋅ (ii)

SquaringEq.(i),wegeta2r2(1+r4)2=25

2⋅⋅⋅⋅⋅⋅ ⋅ (iii)

DivideEq.(ii)by(iii),

⇒a2r2(1+r4)2

a2r6=(25)2

4(25)

⇒(1+r4)2

r4=25

⇒4(1+r4)2=25r4

⇒4(1+r8+2r4) = 25r4

⇒4r8−17r4+4=0

⇒4r8−16r4−r4+4=0

⇒4r4(r4−4) − 1(r4+ (−4)) = 0

⇒ (r4−4)(4r4−1) = 0

Gives, r4=40rr4=1∕4

Wehavetofindsumof4th,6thand8thterm,i.e.

T4+T6+T8=ar3+ar5+ar7

=ar(r2+r4+r6)

=ar3(1+r2+r4) ⋅⋅⋅⋅⋅⋅ ⋅ (iv)

UsingEq.(ii),

(ar3)2=25

⇒ar3=5

Also,wetaker4=4becausegivenseriesisincreasingandr2=2.

∴T4+T6+T8=5(1+2+4)

=5(7) = 35

( )

Wealreadyknow,Arithmeticmean≥Geometricmean,

-------------------------------------------------------------------------------------------------

Question108

If0 <θ,φ<π

2,x=∞

∑

n=0cos2nθ1y =∞

∑

n=0sin2nφand

z=∞

∑

n=0cos2nθ⋅sin2nφ,then

[2021,25Feb.Shift-1]

Options:

A.xy −z= (x+y)z

B.xy +yz +zx =z

C.xyz =4

D.xy +z= (x+y)z

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

LetustakeAMandGMoftwotermsaa*anda1−ax,

⇒AM =aax+a1−ax

and GM =aax⋅a1−ax

∵AM ≥GM ⇒aax+a1−ax

2≥aax⋅a1−ax

⇒aax+a1−ax≥2a1

∴Minimumvalueof f (x) = aax+a1−axis

2√a.

√

Given,x=

∞

∑

n=0

cos2nθ

∞

∑

n=0

sin2nφ

∞

∑

n=0

cos2nθ⋅sin2nφ

⇒x=1+cos2θ+cos4θ+ .. . ∞

∴x=1

1−cos2θ=cosec2θ

⇒y=1+sin2φ+sin4φ+ .. . ∞

∴x=1

1−cos2θ=cosec2θ⋅⋅⋅⋅⋅⋅ ⋅ (i)

⇒y=1+sin2φ+sin4φ+ .. . ∞

∴y=1

1−sin2φ = sec2φ⋅⋅⋅⋅⋅⋅ ⋅ (ii)

⇒z=1+cos2θ⋅sin2φ+cos4θsin4φ + .. . ∞

∴z=1

1−cos2θsin2φ...(iii)

FromEqs.(i),(ii)and(iii),weget

z=1

1−1−1

x1−1

∵cos2θ=1−1

∵sin2φ=1−1



z=xy

xy − (x−1)(y−1)

z=xy

xy −xy +x+y−1

⇒xz +yz −z=xy

⇒xy +z= (x+y)z

( ) ( ) [ ]

Question109

LetA1,A2,A3, ........besquares,suchthatforeachn ≥1,thelengthofthe

sideofAnequalsthelengthofdiagonalofAn+1.IfthelengthofA1is12cm,

thenthesmallestvalueofnforwhichareaofAnislessthanone,is..........

[2021,25Feb.Shift-I]

Answer:9

Solution:

-------------------------------------------------------------------------------------------------

Question110

Consideranarithmeticseriesandageometricserieshavingfourinitialterms

fromtheset{11,8,21,16,26,32,4}.Ifthelasttermsoftheseseriesare

themaximumpossiblefourdigitnumbers,thenthenumberofcommon

termsinthesetwoseriesisequalto.........

[2021,16MarchShift-1]

Accordingtothequestion,lengthofsideofA1squareis12 mathrm ~cm.

∵SidelengthsareinGP.

∴Tn=12

(√2)n−1

(Sideofnthsquarei.e.An)

∴Area = ( Side )2=12

(√2)n−1

2=144

2n−1

Accordingtothequestion,theareaofAnsquare<1

⇒2n−1>144

Here,thesmallestpossiblevalueofis = 9.

( )

Answer:3

Solution:

-------------------------------------------------------------------------------------------------

Question111

Let 1

16,aandbbeinG.P.and 1

a,1

b,6bein(a)P.,wherea,b>0.Then,

72(a+b)isequalto.........

[2021,16MarchShift-II]

Answer:14

Solution:

-------------------------------------------------------------------------------------------------

Question112

Ifα,βarenaturalnumbers,suchthat100α−199β = (100)(100) + (99)(101)+

(98)(102) + ... + (1)(199),thentheslopeofthelinepassingthrough(α,β)

andoriginis

[2021,18MarchShift-1]

Options:

Given,set{11,8,21,16,26,32,4}

Byobservation,wecansaythat

AP = {11,16,21,26, ...}

GP = {4,8,16,32, ...}

5m +6=4.2n−1

5m +6=2n+1

So,(2n+1−6)shouldbeamultipleof5.Theunitdigitof2kis2,4,6,8.So,when6issubtractedfrom2n+1,thepossibleunit

digitswillbe6,8,0,2.Only0isdivisibleby5.Hence,2n+1unitdigithastobe6.

2n+1=24,28,212,216...

As,216willnotbea4digitnumber,so,commonterms = {16,256,4096}

∴Numberofcommonterms = 3

Given, GP =1

16,a,b

⇒a2=b

andgiven,AP =1∕a,1∕b,6

⇒2

b=1

a+6

⇒2

16a2=1

a+6

⇒1

8a2=1+6a

⇒1=8a(1+6a)

⇒48a2+8a −1=0

⇒ (4a +1)(12a −1) = 0

⇒a= −1∕4 or 1 ∕12

Asperthequestion,a>0

∴a=1∕12

b=16a2=16⋅1

144 =1

∴72(a+b) = 72 1

12 +1

=6+8

=14

( )

A.540

B.550

C.530

D.510

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question113

32−1+1

52−1+1

72−1+ ... + 1

(201)2−1

isequalto

[2021,18MarchShift-1]

Options:

A. 101

404

B. 25

101

C. 101

408

D. 99

400

Answer:B

Solution:

Given,

100α−199 .β= (100)(100) + (99)(101)

+(98)(102) + ... + (1)(199)

⇒100α−199β =

∑

x=0

(100 −x)(100 +x)

∑

x=0

(1002−x2)

∑

x=0

(100)2−

∑

x=0

(x)2

= (100)3−99 ×100 ×199

⇒(100)α− (199)β= (100)3− (199)(1650)

Oncomparing,wegetα=3,β=1650

Then,theslopeofthelinepassing

through(α,β)andoriginis

=β−0

α−0=β

α=1650

3=550

32−1+1

52−1+1

72−1+ ... + 1

(201)2−1

100

∑

r=1

(2r +1)2−1

100

∑

r=1

4r2+4r +1−1

100

∑

r=1

2r(2r +2)=

100

∑

r=1

4(r)(r+1)

-------------------------------------------------------------------------------------------------

Question114

LetS1bethesumoffirst2ntermsofanarithmeticprogression.LetS2bethe

sumoffirst4ntermsofthesamearithmeticprogression.If(S2−S1)is1000,

thenthesumofthefirst6ntermsofthearithmeticprogressionisequalto

[2021,18MarchShift-11]

Options:

A.1000

B.7000

C.5000

D.3000

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question115

Iflog32,log3(2x−5), log32x−7

2areinanarithmeticprogression,thenthe

valueofxisequalto

[2021,27JulyShift-1]

Answer:3

Solution:

( )

100

∑

r=1

r−1

r+1

1−1

2+1

2−1

3+1

3−1

4+... + 1

100 −1

101

41−1

101 =1

4×100

101

=25

101

( )

[ ( ) ( ) ( ) ( ) ]

[ ]

Given,S1=S2nandS2=S4n

andS2−S1=1000 ⇒S4n −S2n =1000

⇒4n

2[2a + (4n −1)d] − 2n

2[2a + (2n −1)d] = 1000

⇒2n[2a + (4n −1)d] − n[2a + (2n −1)d] = 1000

⇒2an +n(8n −2−2n +1)d=1000

⇒2an +n(6n −1)d=1000

⇒n[2a + (6n −1)d] = 1000

⇒S6n =6n

2[2a + (6n −1)d]

2× (1000)

=6×500 =3000

Question116

Iftan π

9,x,tan 7π

18 areinarithmeticprogressionandtan π

9,y,tan 5π

arealsoinarithmeticprogression,then|x−2y|isequalto

[2021,27JulyShift-II]

Options:

A.4

B.3

C.0

D.1

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question117

LetSnbethesumofthefirstntermsofanarithmeticprogression,If

S3n =3S2n,thenthevalueof S4n

S2n

is

[2021,25JulyShift-1]

( ) ( ) ( ) ( )

log32,log3(2x−5), log32x−7

2→AP

⇒2log3(2x−5) = log32+log32x−7

⇒log3(2x−5)2=log32⋅2x−7

⇒ (2x−5)2=2⋅2x−7

⇒ (2x)2+25 −10 ⋅2x−2⋅2x+7=0

⇒ (2x)2−12 ⋅2x+32 =0

⇒ (2x−4)(2x−8) = 0

⇒2x=4or8⇒x=2or3

Ifx=2,thenlog3(2x−5) = log3(22−5)

Here,argumentisnegative,so,x≠2.

Hence,x=3

( )

[ ( ) ]

Iftan π

9,x,tan 7π

18 areinAP.

So,x=1

2tan π

9+tan 7π

( ∵ifa,b,careinAP,so,b=a+c

Andtan π

9,y1tan 5π

18 areinAP.

Now,x−2y =1

2tan π

9+tan 7π

18 − tan π

9+tan 5π

18 

⇒ | x−2y | =

cot π

9−tan π

2−tan 5π

18 

∵tan 5π

18 =cot 2π

and tan 7π

18 =cot π

=cot 2π

9−cot 2π

9=0

∵cot 2 A =2cot2A−1

2 cot A

( ) ( )

[ ( ) ]

)

( ) ( )

[ ] ( )

| | { }

| |

[ ]

Options:

A.6

B.4

C.2

D.8

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question118

LetSndenotethesumoffirstntermsofanarithmeticprogression.If

S10 =530,S5=140,thenS20 −S6isequalto

[2021,22JulyShift-II]

Options:

A.1862

B.1842

C.1852

D.1872

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question119

Thesumofalltheelementsintheset{n∈ {1,2, .. . 100}:HCFofnand

2040is1}isequalto

[2021,22JulyShift-II]

LetSn=An2+Bn =n(An +B)S3n =3S2n

⇒3n[A(3n) + B] = 3⋅2n ⋅ [A(2n) + B]

⇒3An +B=4An +2B

⇒An +B=0

∴S4n

S2n

=4n[A(4n) + B]

2n[A(2n) + B]

=24An −An

2An −An

=2×3=6

( )

Sn=An2+Bn

S10 =100A +10B =530

S5=25A +5B =140Solvingbothequations,wegetB=3and

A=5

∴Sn=5n2+3n

∴S20 −S6=5(202−62) + 3(20 −6)

=5⋅26.14 +3⋅14

=14(130 +3) = 14 ×133 =1862

Answer:1251

Solution:

-------------------------------------------------------------------------------------------------

Question120

Ifsumofthefirst21termsoftheseries

log91∕2x+log91∕3x+log91∕4x+ .....wherex >0

is504,thenxisequalto

[2021,20JulyShift-II]

Options:

A.243

B.9

C.7

D.81

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question121

Fork ∈N ,let

α(α+1)(α+2)... . (α+20)=20

∑

k=0

α+k

whereα >0.Thenthevalueof100 A14 +A15

A13

[2021,20JulyShift-II]

( )

n∈ {1,2,3, ..... . 100}

2040 =23×3×5×17

IfHCFofnand2040is1,nshouldnotbeamultipleof2,3,5,17.

n∈ { 1,7,11,13,19,23,29,31,37,41,43,47,53,59,61,67,71,73,77,79,83,89,91,97 }

∑n= | 1251|

LetS=log91∕2x+log91∕3x+ ...

Usingproperty,logabx=1

blogax

S=2log9x+3log9x+ ... + 22log9x

=log9x(2+3+4+ ... + 22)

=log9x21

2(4+20) = log9x(21 ×12)

∴S=252log9x

Given, S=504,then

252log9x=504

⇒log9x=2

⇒x= (9)2=81

[ ]

Answer:9

Solution:

-------------------------------------------------------------------------------------------------

Question122

Ifthevalueof

1+2

3+6

32+10

+.... upto ∞

log(0.25)

3+1

32+1

33+ ...... upto ∞

isl ,thenl 2isequalto.......

[2021,25JulyShift-I]

Answer:3

Solution:

( ) ( )

Given,

α(α+1)... + (α+20)=

∑

k=0

α+k,α>0

⇒1

α(α+1).. . (α+20)=A0

α+A1

α+1

+... + A20

α+20

A14 =1

(−14)(−13).. . (−1)(1).. . (6)=1

14!6!

A15 =1

(−15)(−14).. . (−1)(1).. . (5)=−1

15!5!

A13 =1

(−13).. . (−1)(1).. . (7)=−1

13!7!

A14

A13

=−13!7!

14!6!=−7

14 =−1

⇒A15

A13

=13!7!

15!5!=42

15 ×14 =1

∴100 A14

A13

+A15

A13

=100 −1

2+1

2

=100 −3

2=9

( ) ( )

( )

Let l=αβ

α=1+2

3+6

32+10

33+ .. . ∞⋅⋅⋅⋅⋅⋅ ⋅ (i)

3=1

3+2

32+6

33+ .... ⋅⋅⋅⋅⋅⋅ ⋅ (ii)

SubtractingEq.(ii)fromEq.(i),

2α

3=1+1

3+4

32+4

33+ .....

1−1

∴α=3

β=log0.25

3+1

32+ ....

=log0.25

1−1

=log 1

2=1

∴L=31∕2

( )

-------------------------------------------------------------------------------------------------

Question123

Let{an}n=1

∞beasequencesuch

thata1=1,a2=1andan+2=2an+1+anforalln ≥1.Thenthevalueof

47 ∞

∑

n=1

23nisequalto....

[2021,20JulyShift-II]

Answer:7

Solution:

-------------------------------------------------------------------------------------------------

Question124

Leta1,a2,a3......bean(a)P.

⇒L2=3

Let

∞

∑

n=1

23n =xi.e.

∞

∑

n=1

8n=x

Given,an+2=2an+1+an

Dividethewholeby8n,

an+2

8n=2an+1

8n+an

⇒82⋅an+2

8n+2=8⋅2an+1

8n+1+an

8n⇒

64 an+2

8n+2=16 an+1

8n+1+an

Now,takethesummation,

∞

∑

n=1

an+2

8n+2=16

∞

∑

n=1

an+1

8n+1+

∞

∑

n=1

8n⋅⋅⋅⋅⋅⋅ ⋅ (i)

∵

∞

∑

n=1

8n=x

i.e. a1

8+a2

82+a3

83+a4

84+ ... = x

⇒a3

83+a4

84+ ... = x−a1

8−a2

⇒

∞

∑

n=1

an+2

8n+2=x−a1

8−a2

82⋅⋅⋅⋅⋅⋅ ⋅ (ii)

Again, a2

82+a3

83+ ... = x−a1

⇒

∞

∑

n=1

an+1

8n+1=x−a1

8⋅⋅⋅⋅⋅⋅ ⋅ (iii)

FromEqs.(i),(ii)and(iii),

64 x −a1

8−a2

64 =16 x −a1

8+x

Usea1=1=a2

64 x −1

8−1

64 =16 x −1

8+x

⇒64x −9=2(8x −1) + x

⇒64x −16x −x=9−2⇒47x =7

∴47 ∑an

23n =7

( ) ( )

If a1+a2+ ... + a10

a1+a2+ ..... + ap

=100

p2,p≠10,then a11

a10

isequalto

[2021,31Auq.Shift-II]

Options:

A. 19

B. 100

121

C. 21

D. 121

100

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question125

Thenumberof4-digitnumberswhichareneithermultipleof7normultiple

of3is

[2021,31Aug.Shift-II]

Answer:5143

Solution:

a1+a2+ ... + a10

a1+a2+ ... + ap

=100

⇒S10

=100

p2⇒Sp=S10 ⋅p2

100

a11

a10

=S11 −S10

S10 −S9

S10 ⋅121

100 −S10

S10 −S10⋅81

100

121

100 −1

1−81

100

=21

Total4-digitnumber

=9×10 ×10 ×10 =9000

4-digitnumberdivisibleby7

1001,1008, ..., 9996

Numberof4-digitnumberdivisibleby7

=9996 −1001

7+1=1286

4-digitnumberdivisibleby3

1002,1005, ..., 9999

Numberof4-digitnumberdivisibleby3

=9999 −1002

3+1=3000

4digitnumberdivisibleby21

1008,1031, ..., 9996

Numberof4-digitnumberdivisibleby21

=9996 −1008

21 +1=429

∴Numberof4-digitnumbersneitherdivisibleby7nor3

=9000 −1286 −3000 +429 =5143

Question126

Threenumbersareinanincreasinggeometricprogressionwithcommon

ratior.Ifthemiddlenumberisdoubled,thenthenewnumbersareinan

arithmeticprogressionwithcommondifferenced .IfthefourthtermofGPis

3r2,thenr2−d isequalto

[2021,31Aug.Shift-I]

Options:

A.7 −7√3

B.7 + √3

C.7 − √3

D.7 +3√3

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question127

If0 <x<1,then

2x2+5

3x3+7

4x4+ ......,isequalto

[2021,27Aug.Shift-1]

Options:

A.x 1+x

1−x+loge(1−x)

B.x 1−x

1+x+loge(1−x)

C. 1−x

1+x+loge(1−x)

D. 1+x

1−x+loge(1−x)

( )

Letthreenumbersbe a

r,a,ar.

Accordingtothequestion, a

r,2a,ar →AP

4a =ar +a

r⇒r+1

r=4

⇒r2−4r +1=0⇒r=2± √3

T4ofGP =3r2

3r2=ar2

a=3

r=2+ √3

d=2a −a

r=3√3

r2−d= (2+ √3)2−3√3

=7+4√3−3√3=7+ √3

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question128

Ifx,y∈R,x>0

y=log10x+log10x1∕3+log10x1∕9+ ...

upto∞termsand

2+4+6+ ... + 2y

3+6+9+ ... + 3y =4

log10x,thentheorderedpair(x,y)isequalto

[2021,27Aug.Shift-1]

Options:

A.(106,6)

B.(104,6)

C.(102,3)

D.(106,9)

Answer:D

Solution:

Wehave,

2x2+5

3x3+7

4x4+ ...

=2−1

2x2+2−1

3x3+2−1

4x4+ ...

=2(x2+x3+x4+ ...) − x2

2+x3

3+x4

4+ ...

=2⋅x2

1−x− [−loge(1−x) − x]

[usingsumofinfiniteGP =a

1−randlogarithmicseries]

=2x2

1−x+x+loge(1−x)

=2x2+x−x2

1−x+loge(1−x)

=x2+x

1−x+loge(1−x)

=x1+x

1−x+loge(1−x)

( ) ( ) ( )

( )

Given, 2+4+6+ ... + 2y

3+6+9+ ... + 3y =4

log10x

⇒2(1+2+3+ ... + y)

3(1+2+3+ ... + y)=4

log10x

⇒2

3=4

log10x⇒log10x=6

⇒x=106

Now,y=log10x+log10x

3+log10x

9+ ...upto∞terms.

=log10 x⋅x

3⋅x

9...∞ terms 

=log10x

1+1

3+1

9+ .. . ∞ terms

( )

-------------------------------------------------------------------------------------------------

Question129

If0 <x<1andy =1

2x2+2

3x3+3

4x4+ ...thenthevalueofe1+yatx =1

2is

[2021,27Aug.Shift-II]

Options:

A. 1

2e2

B.2e

C. 1

2√e

D.2e2

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question130

IfthesumofaninfiniteGPa,ar,ar2,ar3, ....is15andthesumofthe

squaresofitseachtermis150,thenthesumofar2,ar4,ar6, ...is

[2021,26Aug.Shift-1]

Options:

A.5 ∕2

B.1 ∕2

C.25 ∕2

=log10x

1−1

3=log10x3∕2

=log10(106)

2[∵x=106]

⇒y=6×3

2=9

∴x=106,y=9

(x,y) = (106,9)

y=1

2x2+2

3x3+3

4x4+ ...

⇒y=1−1

2x2+1−1

3x3+1−1

4x4+

= (x2+x3+x4+ ...) − x2

2+x3

3+x4

4+ ...

=x2

1−x+x−x+x2

2+x3

3+x4

4...

∴y=x

1−x+ln(1−x)

Putx=1

2,weget

y=1−ln 2

Then,e1+y=e1+1−ln 2 =e2−ln 2

=e2⋅eln 2−1

2e2

( ) ( ) ( )

( )

D.9 ∕2

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question131

Leta1,a2......, a10beanAPwithcommondifference−3andb1,b2, ......, b10

beaGPwithcommonratio2.

Letck=ak+bk,k=1,2, ......, 10.Ifc2=12and

c3=13,then 10

∑

k=1ckisequalto

[2021,26Aug.Shift-II]

Answer:2021

Solution:

Wehave,sumofinfiniteGPa,ar,ar2, ...is

S∞=a

1−r=15 ⋅⋅⋅⋅⋅⋅ ⋅ (i)

andsumofinfiniteGPa2,a2r2,a2r4, ...is

cS∞

′=a2

1−r2=150

⇒a

1−r

1+r=150 ⋅⋅⋅⋅⋅⋅ ⋅ (ii)

DivideEq.(ii)byEq.(i)

1+r=10 ⋅⋅⋅⋅⋅⋅ ⋅ (iii)

DivideEq.(iii)byEq.(i)

1−r

1+r=10

15 =2

⇒3−3r =2+2r

⇒1=5r ⇒r=1

Now,puttingr=1

5inEq.(iii),weget

1+1

=10

⇒5a

6=10 ⇒a=12

Now,sumofar2,ar4,ar6, ..., ∞

S∞

′′ =ar2

1−r2=

12 ⋅1

( ) ( )

( )

a1,a2,a3, ..., a10areinAPcommon

difference = −3

b1,b2,b3, ..., b10areinGPcommonratio = 2

Since,c = ak+bk,k=1,2,3......, 10

∴c2=a2+b2=12

c3=a3+b3=13

Now,c3−c2=1

⇒ (a3−a2) + (b3−b2) ≠ 1

⇒ − 3+ (2b2−b2) ≠ 1

⇒b2=4

∴a2=8

So,APis11,8,5, ....

-------------------------------------------------------------------------------------------------

Question132

Thesumof10termsoftheseries 3

12×22+5

22×32+7

32×42+ ...is

[2021,31Aug.Shift-1]

Options:

A.1

B.120/121

C.99/100

D.143 ∕144

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question133

IfS =7

5+9

52+13

53+19

54+ ...,then160Sisequalto

[2021,31Aug.Shift-II]

Answer:305

Solution:

andGPis2,4,8, ....

Now,

∑

k=1

ck=

∑

k=1

ak+

∑

k=1

=10

2[22 +9(−3)] + 2210 −1

2−1

=5(22 −27) + 2(1023)

=2046 −25 =2021

( ) ( )

12×22+5

22×32+7

32×42+ ...

=22−12

12×22+32−22

22×32+42−32

32×42+ ....

12−1

22+1

22−1

32+1

32−1

42+ .... 1

102−1

112

12−1

112=1−1

121 =120

121

( ) ( ) ( ) ( )

S=7

5+9

52+13

53+19

54+ ... + ∞⋅⋅⋅⋅⋅⋅ ⋅ (i)

5=7

52+9

53+13

54+19

55+ .. . ∞⋅⋅⋅⋅⋅⋅ ⋅ (ii)

SubtractingEq.(ii)fromEq.(i),

c4S

5=7

5+2

52+4

53+6

54+8

55+ .. . ∞

5−7

5=2

52+4

53+6

54+8

55+ .. . ∞=K⋅⋅⋅⋅⋅⋅ ⋅ (iii)

5=2

53+4

54+6

55+8

56+ .. . ∞... (iv)

-------------------------------------------------------------------------------------------------

Question134

Leta1,a2, ......, a21beanAPsuchthat 20

∑

n=1

anan+1

9.IfthesumofthisAPis

189,thena6a16isequalto

[2021,01Sep.Shift-II]

Options:

A.57

B.72

C.48

D.36

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question135

LetSn=1⋅ (n−1) + 2⋅ (n−2) + 3.

(n−3) + ... + (n−1) ⋅ 1,n≥4 .

SubtractingEq.(iv)fromEq(iii),

5=2

52+2

53+2

54+2

55+ .. . ∞

5=2

1−1∕5=1

10 ⇒K=1

FromEq.(iii),

5−7

5=1

8⇒S=61

Now,106S =160×61

32 =305

( )

LetdbethecommondifferenceofanAP

a1,a2, ..., a21 and

∑

n=1

anan+1

⇒

∑

n=1

an(an+d)=4

⇒1

∑

n=1

−1

an+d=4

⇒1

−1

+ ... + 1

a20

−1

a21

⇒1

−1

a21

⇒1

a21 −a1

a1a21

9⇒a1a21 =45

⇒a1(a1+20d ) = 45 ⋅⋅⋅⋅⋅⋅ ⋅ (i)

⇒21

2(2a1+20d ) = 189

Alsosumoffirst21terms = 189

⇒21

2(2a1+20d ) = 189

⇒a1+10d =9⋅⋅⋅⋅⋅⋅ ⋅ (ii)

ByEqs.(i)and(ii),wegeta1=3,d=3∕5

ora1=15,d= − 3

So, a6a16 = (a1+5d )(a1+15d ) = 72

( )

[ ]

( )

Thesum ∞

∑

n=4

2Sn

n!−1

(n−2)! isequalto

[2021,01Sep.Shift-II]

Options:

A. e−1

B. e−2

C. e

D. e

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question136

ThenumberoftermscommontothetwoA.P.'s3,7,11, ...407and

2,9,16, ..., 709is______.

[NAJan.9,2020(II)]

Answer:14

Solution:

-------------------------------------------------------------------------------------------------

Question137

Ifthe10 th termofanA.P.is 1

20andits20 th termis 1

10,thenthesumofits

( )

Sn=1⋅ (n−1) + 2(n−2) + 3(n−3)+... + (n−1) ⋅ 1,n≥4

n−1

∑

r=1

r(n−r) = n(n2−1)

=n(n−1)(n+1)

2Sn

n!=(n+1)

3(n−2)!

⇒

∞

∑

n=4

2sn

n!−1

(n−2)!

∞

∑

n=4

n−2

3(n−2)! =1

∞

∑

n=4

(n−3)!

1!+1

2!+1

3!+ ... = e−1

( )

Firstcommontermofboththeseriesis23andcommondifferenceis7×4=28

∵Lastterm≤407⇒23 + (n−1) × 28 ≤407

⇒(n−1) × 28 ≤384

⇒n≤384

28 +1

⇒n≤14.71

Hence,n=14

first200termsis:

[Jan.8,2020(II)]

Options:

A.50

B.50 1

C.100

D.100 1

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question138

Letf :R→Rbesuchthatforallx ∈R, (21+x+21−x), f(x)and(3x+3−x)are

inA.P.,thentheminimumvalueoff (x)is:

[Jan.8,2020(I)]

Options:

A.2

B.3

C.0

D.4

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question139

FivenumbersareinA.P.,whosesumis25andproductis2520.Ifoneof

thesefivenumbersis−1

2,thenthegreatestnumberamongstthemis:

T10 =1

20 =a+9d ...(i)

T20 =1

10 =a+19d ...(ii)

Solvingequations(i)and(ii),weget

a=1

200,d=1

200

⇒S200 =200

200 +199

200 =201

2=100 1

[ ]

If21−x+21+x,f(x), 3x+3−xareinA.P.,then

f(x) = 21+x+21−x+3x+3−x

2

2f (x) = 2 2x+1

2x+3x+1

3x

UsingAM ≥GM 

f(x) ≥ 3

( )

( ) ( )

[Jan.7,2020(I)]

Options:

A.27

B.7

C. 21

D.16

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question140

Theproduct2

4⋅4

16 ⋅8

48 ⋅16

128 ...to∞isequalto:

[Jan.9,2020(I)]

Options:

A.2

B.2

C.1

D.2

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Let5termsofA.P.be

a−2d ,a−d,a,a+d,a+2d 

Sum =25 ⇒5a =25 ⇒a=5

Product = 2520

(5−2d )(5−d)5(5+d)(5+2d ) = 2520

⇒(25 −4d 2)(25 −d2) = 504

⇒625 −100d 2−25d 2+4d 4=504

⇒4d 4−125d 2+625 −504 =0

⇒4d 4−125d 2+121 =0

⇒4d 4−121d 2−4d 2+121 =0

⇒(d2−1)(4d 2−121) = 0

⇒d= ±1,d= ± 11

2

d= ±1andd= − 11

2,doesnotgive −1

2asaterm

∴d=11

2

∴Largestterm = 5+2d =5+11 =16

4+2

16 +3

48 + ... . ∞



4+1

8+1

16 + ... . ∞

= √2

Question141

Letanbethen th termofaG.P.ofpositiveterms.

If 100

∑

n=1a2n +1=200and 100

∑

n=1a2n =100,then 200

∑

n=1anisequalto:

[Jan.9,2020(II)]

Options:

A.300

B.225

C.175

D.150

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question142

Ifx =∞

∑

n=0(−1)ntan2nθandy =∞

∑

n=0cos2nθ,for0 <θ<π

4,then:

[Jan.9,2020(II)]

Options:

A.x(1+y) = 1

B.y(1−x) = 1

C.y(1+x) = 1

D.x(1−y) = 1

Answer:B

Solution:

LetG.P.bea,ar,ar2......

100

∑

n=1

a2n +1=a3+a5+ ..... + a201 =200

⇒ar2(r200 −1)

r2−1=200

100

∑

n=1

a2n =a2+a4+ ..... + a200 =100

⇒ar(r200 −1)

r2−1=100

Fromequations(i)and(ii),r=2and

a2+a3+ ..... + a200 +a201 =300

⇒r(a1+ ..... + a200) = 300

⇒

200

∑

n=1

an=300

r=150

y=1+cos2θ+cos4θ+ ....

⇒y=1

1−cos2θ⇒1

y=sin2θ

x=1−tan2θ+tan4θ+ ....

x=1

1− (−tan2θ)=1

sec2θ⇒x=cos2θ

-------------------------------------------------------------------------------------------------

Question143

Thegreatestpositiveintegerk,forwhich49k+1isafactorofthesum

49125 +49124 + ... + 492+49 +1,is:

[Jan.7,2020(I)]

Options:

A.32

B.63

C.60

D.65

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question144

Leta1,a2,a3, ...beaG.P.suchthata1<0,a1+a2=4anda3+a4=16.If

∑

i=1ai=4λ,thenλisequalto:

[Jan.7,2020(II)]

Options:

A.-513

B.-171

C.171

D. 511

Answer:B

Solution:

y=1

sin2θ⇒y=1

1−x

∴y(1−x) = 1

(49)126 −1

48 =((49)63 +1)(4963 −1)

48 ∵Sn=a(rn−1)

r−1

∴K=63

[ ]

Since,a1+a2=4⇒a1+a1r=4...(i)

a3+a4=16 ⇒a1r2+a1r3=16...(ii)

Fromeqn.(i),a1=4

1+randsubstitutingthevalueofa1,ineqn(ii),

1+r

=16

⇒4r2(1+r) = 16(1+r)

⇒r2=4∴r= ±2

r=2,a1(1+2) = 4⇒a1=4

3

r= −2,a1(1−2) = 4⇒a1= −4

( ) ( )

-------------------------------------------------------------------------------------------------

Question145

Thecoefficientofx7intheexpression

(1+x)10 +x(1+x)9+x2(1+x)8+ ... + x10is:

[Jan.7,2020(II)]

Options:

A.210

B.330

C.120

D.420

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question146

Thesum, 7

∑

n=1

n(n+1)(2n +1)

4isequalto_________.

[Jan.8,2020(II)]

Answer:504

Solution:

∑

i=1

ai=a1(rq−1)

r−1=(−4)((−2)9−1)

−2−1

3(−513) = 4λ ⇒λ= −171

ThegivenseriesisinG.P.thenSn=a(1−rn)

1−r

(1+x)10 1−x

1+x

1−x

1+x



⇒(1+x)10[(1+x)11 −x11]

(1+x)11×1

(1+x)

= (1+x)11 −x11

∴ Coefficientofx7is11C7=11C11 −7=11C4=330

[ ( ) ]

( )

∑

n=1

n(n+1)(2n +1)

∑

n=1

(2n3+3n2+n) 

427.8

2+37.8.15

6+7.8

2

⇒1

4[2×49 ×16 +28 ×15 +28]

4[1568 +420 +28] = 504

[ ] [ ]

[ ( ) ( ) ]

Question147

Thesum 20

∑

k=1(1+2+3+ ... + k)is

[Jan.8,2020(I)]

Answer:1540

Solution:

-------------------------------------------------------------------------------------------------

Question148

Ifthesumofthefirst40termsoftheseries,3 +4+8+9

+13 +14 +18 +19 + ...is(102)m,thenmisequalto:

[Jan.7,2020(II)]

Options:

A.20

B.25

C.5

D.10

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question149

Ifthesumoffirst11termsofanA.P.,a1,a2,a3, ...is0(a1≠0),thenthesum

oftheA.P.,a1,a3,a5, ..., a23iska1,wherekisequalto:

[Sep.02,2020(II)]

Options:

A.−121

Givenseriescanbewrittenas

∑

k=1

k(k+1)

2=1

∑

k=1

(k2+k)

20(21)(41)

6+20(21)

2

420 ×41

6+20 ×21

2=1

2[2870 +210] = 1540

[ ]

S=3+4+8+9+13 +14 +18 +19..... . 40terms

S=7+17 +27 +37 +47 + ... . .20terms

S40 =20

2[2×7+ (19)10] = 10[14 +190]

=10[2040] = (102)(20)

⇒m=20

B. 121

C. 72

D.−72

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question150

IfthefirsttermofanA.P.is3andthesumofitsfirst25termsisequaltothe

sumofitsnext15terms,thenthecommondifferenceofthisA.P.is:

[Sep.03,2020(I)]

Options:

A. 1

B. 1

C. 1

D. 1

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question151

Inthesumoftheseries20 +19 3

5+19 1

5+18 4

5+ ...upton th termis488and

Letcommondifferencebed

∵S11 =0∴11

2{2a1+10 ⋅d} = 0

⇒a1+5d =0⇒d= − a1

5...(i)

Now,S=a1+a3+a5+ ... + a23

=a1+ (a1+2d ) + (a1+4d ) + .... + (a1+22d )

=12a1+2d 11 ×12

2

=12 a1+11 ⋅ − a1

5(From(i))

=12 × − 6

5a1= − 72

5a1

[ ( ) ]

( )

Givena=3andS25 =S40 −S25

⇒2S25 =S40

⇒2×25

2[6+24d ] = 40

2[6+39d ]

⇒25[6+24d ] = 20[6+39d ]

⇒5(2+8d ) = 4(2+13d )

⇒10 +40d =8+52d 

⇒d=1

thenn th termisnegative,then:

[Sep.03,2020(II)]

Options:

A.n =60

B.n th termis-4

C.n =41

D.n th termis−42

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question152

Leta1,a2, ...., anbeagivenA.P.whosecommondifferenceisanintegerand

Sn=a1+a2+ .... + an.Ifa1=1,an=300and15 ≤n≤50,thentheordered

pair(Sn−4,an−4)isequalto:

[Sep.04,2020(II)]

Options:

A.(2490,249)

B.(2480,249)

C.(2480,248)

D.(2490,248)

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Sn=20 +19 3

5+19 1

5+18 4

5+ ....

∵Sn=488

488 =n

22100

5+ (n−1) − 2

5488 =n

2(101 −n) ⇒ n2−101n +2440 =0

⇒n=61or40

Forn=40 ⇒Tn>0

Forn=61 ⇒Tn<0

nth term = T61 =100

5+ (61 −1) − 2

5= −4

[ ( ) ( ) ]

( )

Giventhata1=1andan=300andd∈Z

∴300 =1+ (n−1)d

⇒d=299

(n−1)=23 ×13

(n−1)

∵disaninteger

∴n−1=13or23

⇒n=14or24 (∵15 ≤n≤50)

⇒n=24andd=13

a20 =1+19 ×13 =248

s20 =20

2(2+19 ×13) = 2490

Question153

If32 sin 2 α −1,14and34−2 sin 2 αarethefirstthreetermsofanA.P.forsomeα,

thenthesixthtermofthisA.Pis:

[Sep.05,2020(I)]

Options:

A.66

B.81

C.65

D.78

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question154

Ifthesumofthefirst20termsoftheseries

log(71∕2)x+log(71∕3)x+log(71∕4)x+ ...is460,thenxisequalto:

[Sep.05,2020(II)]

Options:

A.72

B.7

C.e2

D.7

Answer:A

Solution:

Giventhat32 sin 2 α −1,14,34−2 sin 2 αareinA.P.

So,32 sin 2 α −1+34−2 sin 2 α =28

⇒32 sin 2 α

3+81

32 sin 2 α =28

Let32 sin 2 α =x

⇒x

3+81

x=28

⇒x2−84x +243 =0⇒x=81,x=3

Whenx=81 ⇒sin 2 α =2(Notpossible)

Whenx=3⇒α=π

12

∴a=30=1,d=14 −1=13

a6=a+5d =1+65 =66

S=log7x2+log7x3+log7x4+ .. . 20terms

∵S=460

⇒log7(x2⋅x3⋅x4⋅..... . x21) = 460

⇒log7x(2+3+4..... . 21)=460

⇒(2+3+4+ ..... + 21)log7x=460

-------------------------------------------------------------------------------------------------

Question155

Iff (x+y) = f(x)f(y)and ∞

∑

x=1f(x) = 2,x,y∈N,whereN isthesetofallnatural

numbers,thenthevalueof f(4)

f(2)is:

[Sep.06,2020(I)]

Options:

A. 2

B. 1

C. 1

D. 4

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question156

Leta,b,c,d andpbeanynonzerodistinctrealnumberssuchthat

(a2+b2+c2)p2−2(ab +bc +cd )p+ ( b2+c2+d 2) = 0.Then:

[Sep.06,2020(I)]

Options:

A.a,c,pareinA.P.

B.a,c,pareinG.P.

C.a,b,c,d areinG.P.

D.a,b,c,d areinA.P.

Answer:C

Solution:

⇒20

2(2+21)log7x=460

⇒log7x=460

230 =2⇒x=72=49

Letf (1) = k,thenf(2) = f(1+1) = k2

f(3) = f(2+1) = k3

∞

∑

x=1

f(x) = 2⇒k+k2+k3+ ... . ∞=2

⇒k

1−k=2⇒k=2

3

Now, f(4)

f(2)=k4

k2=k2=4

Rearrangegivenequation,weget

(a2p2−2abp +b2) + (b2p2−2bcp +c2)

+(c2p2−2cd p +d2) = 0

-------------------------------------------------------------------------------------------------

Question157

ThecommondifferenceoftheA.P.b1,b2, ..., bmis2morethanthecommon

differenceofA.P.a1,a2, ..., an.Ifa40 = −159,a100 = −399andb100 =a70,then

b1isequalto:

[Sep.06,2020(II)]

Options:

A.81

B.-127

C.-81

D.127

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question158

Supposethatafunctionf :R→Rsatisfiesf (x+y) = f(x)f(y)forallx,y∈R

andf (a) = 3.If ∑

i=1

nf(i) = 363,thennisequalto_______.

[NASep.06,2020(II)]

Answer:5

Solution:

⇒(ap −b)2+ (bp −c)2+ (cp −d)2=0

∴ap −b=bp −c=cp −d=0

⇒b

a=c

b=d

c∴a,b,c,dareinG.P.

Letcommondifferenceofseries

a1,a2,a3, ......, anbed

∵a40 =a1+39d = −159...(i)

anda100 =a1+99d = −399...(ii)

Fromequations(i)and(ii),

d= −4anda1= −3

Since,thecommondifferenceofb1,b2, ......, bnis2morethancommondifferenceofa1,a2, ......, an

∴Commondifferenceofb1,b2,b3, .....is(-2)

∵b100 =a70

⇒b1+99(−2) = (−3) + 69(−4)

⇒b1=198 −279 ⇒b1= −81

∵f(x+y) = f(x) ⋅ f(y) ∀x∈Randf(1) = 3

⇒f(x) = 3x⇒f(i) = 3i

⇒

∑

i=1

f(i) = 363

⇒3+32+33+ .... + 3n=363

-------------------------------------------------------------------------------------------------

Question159

If210 +29⋅31+28⋅32+ .... + 2×39+310 =S−211thenSisequalto:

[Sep.05,2020(I)]

Options:

A.311 −212

B.311

C. 311

2+210

D.2 ⋅311

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question160

Ifthesumofthesecond,thirdandfourthtermsofapositivetermG.P.is3

andthesumofitssixth,seventhandeighthtermsis243,thenthesumofthe

first50termsofthisG.P.is:

[Sep.05,2020(II)]

Options:

A. 1

26(349 −1)

B. 1

26(350 −1)

C. 2

13(350 −1)

D. 1

13(350 −1)

Answer:B

Solution:

⇒3(3n−1)

3−1=363 ∵Sn=a(rn−1)

(r−1)

⇒3n−1=363 ×2

3=242

⇒3n=243 =35⇒n=5

[ ]

GivensequenceareinG.P.andcommonratio3

∴

210 3

11 −1

2−1

=S−211

⇒210

311 −211

211

=S−211

⇒311 −211 =S−211 ⇒S=311

( ( ) )

( )

-------------------------------------------------------------------------------------------------

Question161

Letαandβbetherootsofx2−3x +p=0andγandδbetherootsof

x2−6x +q=0.Ifα,β,γ,δformageometricprogression.Thenratio

(2q +p) : (2q −p)is:

[Sep.04,2020(I)]

Options:

A.3:1

B.9:7

C.5:3

D.33:31

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question162

Thevalueof(0.16)log2.5

3+1

32+1

33+ ...to∞ )isequalto________.

[NASep.03,2020(I)]

Answer:4

Solution:

(

Letthefirsttermbe'a'andcommonratiobe'r'.

∵ar(1+r+r2) = 3...(i)

andar5(1+r+r2) = 243...(ii)

From(i)and(ii),

r4=81 ⇒r=3anda=1

13

∴S50 =a(r50 −1)

r−1=350 −1

26 ∵Sn=a(rn−1)

(r−1)

[ ]

Letα,β,γ,δbeinG.P.,thenαδ =βγ

⇒α

β=γ

δ⇒α−β

α+β= ∣ γ−δ

γ+δ

⇒√9−4p

3=√36 −4q

6

⇒36 −16p =36 −4q ⇒q=4p

∴2q +p

2q −p=8p +p

8p −p=9p

7p =9

| |

(0.16)

log2.5

3+1

32+1

33+ .... . ∞



0.16

log2.5

1−1

3∵S∞=a

1−r

( )

( ) [ ]

-------------------------------------------------------------------------------------------------

Question163

ThesumofthefirstthreetermsofaG.P.isSandtheirproductis27.Then

allsuchSliein:

[Sep.02,2020(I)]

Options:

A.(−∞, −9] ∪ [3,∞)

B.[−3,∞)

C.(−∞, −3] ∪ [9,∞)

D.(−∞,9]

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question164

If|x| < 1, | y| <1andx ≠y,thenthesumtoinfinityofthefollowingseries

(x+y) + (x2+xy +y2) + (x3+x2y+xy2+y3) + ....is:

[Sep.02,2020(I)]

Options:

A. x+y−xy

(1+x)(1+y)

B. x+y+xy

(1+x)(1+y)

C. x+y−xy

(1−x)(1−y)

D. x+y+xy

(1−x)(1−y)

Answer:C

=0.16

log2.5

2

= (2.5)−2log2.5

2=1

−2=4

( )

( ) ( )

LettermsofG.P.be a

r,a,ar

∴a1

r+1+r=S...(i)

anda3=27

⇒a=3...(ii)

Puta=3ineqn.(1),weget

S=3+3 r +1

r

Iff(x) = x+1

x,thenf(x) ∈ (−∞, −2] ∪ [2,∞)

⇒3f (x) ∈ (−∞, −6] ∪ [6,∞)

⇒3+3f (x) ∈ (−∞, −3] ∪ [9,∞)

Then,itconcludesthat

S∈ (−∞, −3] ∪ [9,∞)

( )

Solution:

-------------------------------------------------------------------------------------------------

Question165

LetSbethesumofthefirst9termsoftheseries:

{x+ka} + {x2+ (k+2)a} + {x3+ (k+4)a}+{x4+ (k+6)a} + ... where a ≠0

andx ≠1.IfS =x10 −x+45a(x−1)

x−1,thenkisequalto:

[Sep.02,2020(II)]

Options:

A.-5

B.1

C.-3

D.3

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question166

Ifmarithmeticmeans(A.Ms)andthreegeometricmeans(G.Ms)areinserted

between3and243suchthat4 th A.M.isequalto2 nd G.M.,thenmisequal

to_______

[Sep.03,2020(II)]

Answer:39

Solution:

S= (x+y) + (x2+y2+xy) + (x3+x2y+xy2+y3) + .. . ∞

x−y[ (x2−y2) + (x3−y3) + (x4−y4+ .. . ∞]

x−y

1−x−y2

1−y=(x−y)(x+y−xy)

(x−y)(1−x)(1−y)

∵S∞=a

1−r

=x+y−xy

(1−x)(1−y)

[ ]

S= ( x+x2+x3+ .. . 9terms)

+a[k+ (k+2)+ + (k+4) + .. . 9terms]⇒S=x(x9−1)

x−1+9

2[2ak +8× (2a)]

⇒S=x10 −x

x−1+9a(k+8)

1=x10 −x+45a(x−1)

x−1Given)

⇒x10 −x+9a(k+8)(x−1)

x−1=x10 −x+45a(x−1)

x−1

⇒9a(k+8) = 45a ⇒k+8=5⇒k= −3

(

LetmarithmeticmeanbeA1,A2...AmandG1,G2,G3begeometricmean.

-------------------------------------------------------------------------------------------------

Question167

If1 + (1−22⋅1) + (1−42⋅3) + (1−62⋅5) + ...... + (1−202)19) =α−220β,

thenanorderedpair(α,β)isequalto:

[Sep.04,2020(I)]

Options:

A.(10,97)

B.(11,103)

C.(10,103)

D.(11,97)

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question168

Letf :R→Rbeafunctionwhichsatisfiesf (x+y) = f(x) + f(y), ∀x,y∈R.If

f(a) = 2andg(n) = (n−1)

∑

k=1f(k), n∈N,thenthevalueofn,forwhichg(n) = 20,

is:

[Sep.02,2020(II)]

Options:

A.5

B.20

TheA.P.formedbyarithmeticmeanis,

3,A1,A2,A3, ..... . Am,243

∴d=243 −3

m+1=240

m+1

TheG.P.formedbygeometricmean

3,G1,G2,G3,243

r=243

3+1= (81)1∕4=3

∵A4=G2

⇒3+4240

m+1=3(3)2

⇒3+960

m+1=27 ⇒m+1=40 ⇒m=39.

( )

Thegivenseriesis

1+ (1−22⋅1) + (1−42⋅3) + (1−62⋅5) + .. . (1−202⋅19)

S=1+

∑

r=1

[1− (2r)2(2r −1)]

=1+

∑

r=1

(1−8r3+4r2) = 1+10 −

∑

r=1

(8r3−4r2)

=11 −810 ×11

2+4×10 ×11 ×21

6

=11 −2× (110)2+4×55 ×7

=11 −220(110 −7)

=11 −220 ×103 =α−220β

⇒α=11,β=103

∴(α,β) = (11,103)

( ) ( )

C.4

D.9

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question169

Leta,bandcbethe7 th ,11 th and13 th termsrespectivelyofanon-

constantA.P.IfthesearealsothethreeconsecutivetermsofaG.P.,then a

cis

equalto:

[Jan.09,2019(II)]

Options:

A.2

B. 1

C. 7

D.4

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question170

Ifa,bandcbethreedistinctrealnumbersinG.P.anda +b+c=xb,thenx

Given:f(x+y) = f(x) + f(y), ∀x,y∈R,f(1) = 2

⇒f(2) = f(1) + f(1) = 2+2=4

f(3) = f(1) + f(2) = 2+4=6

f(n−1) = 2(n−1)

Now,g(n) =

n−1

∑

k=1

f(k)

=f(1) + f(2) + f(3) + ... . f(n−1)

=2+4+6+ ..... + 2(n−1)

=2[1+2+3+ ...... + (n−1)]

=2×(n−1)(n)

2=n2−n

∵g(n) = 20(given)

So,n2−n=20

⇒n2−n−20 =0

⇒(n−5)(n+4) = 0

⇒n=5orn= −4(notpossible)

LetfirsttermandcommondifferencebeAandDrespectively.

∴a=A+6D,b=A+10Dandc=A+12D

Since,a,b,careinG.P.

Hence,relationbetweena,bandcis,

∴b2=a.c.

∴(A+10D)2= (A+6D)(A+12D)

∴14D +A=0

∴A= −14D

∴a= −8D,b= −4Dandc= −2D

∴a

c=−8D

−2D =4

cannotbe:

[Jan.09,2019(I)]

Options:

A.-2

B.-3

C.4

D.2

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question171

Thesumofthefollowingseries1 +6+9(12+22+32)

7+12(12+22+32+42)

+15(12+22+ ... + 52)

11 + ...upto15terms,is:

[Jan.09,2019(II)]

Options:

A.7520

B.7510

C.7830

D.7820

Answer:D

Solution:

∵a,b,c,areinG.P.

⇒b2=ac

Since,a+b+c=xb

⇒a+c= (x−1)b

Takesquareonbothsides,weget

a2+c2+2ac = (x−1)2b2

⇒a2+c2= (x−1)2ac −2ac[∵b2=ac]

⇒a2+c2=ac[(x−1)2−2]

⇒a2+c2=ac[x2−2x −1]

∵a2+c2arepositiveandb2=acwhichisalsopositive.

Then,x2−2x −1wouldbepositivebutforx=2,x2−2x −1isnegative.

Hence,xcannotbetakenas2.

S=1+6+9(12+22+32)

7+12(12+22+32+42)

9+ 15(12+22+32+42+52)

11 + ...

S=3⋅ (1)2

3+6⋅ (12+22)

5+9⋅ (12+22+32)

7+ 12 ⋅ (12+22+32+42)

9+ ...

Now,nth termoftheseries,

tn=3n ⋅ (12+22+ ... + n2)

(2n +1)

⇒tn=3n ⋅n(n+1)(2n +1)

6(2n +1)=n3+n2

2

∴Sn=Σtn=1

n(n+1)

2+n(n+1)(2n +1)

6

=n(n+1)

n(n+1)

2+2n +1

3

{ ( ) }

( )

-------------------------------------------------------------------------------------------------

Question172

Thesumofallvaluesofθ ∈0,π

2satisfyingsin22θ +cos42θ =3

4is:

[Jan.10,2019(I)]

Options:

A.π

B. 5π

C. π

D. 3π

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question173

Letαandβbetherootsofthequadraticequation

x2sin θ −x(sin θ cos θ +1) + cos θ =0(0<θ<45∘),andα <β.Then

∑

n=0

∞αn+(−1)n

βnisequalto:

[Jan.11,2019(II)]

Options:

A. 1

1−cos θ −1

1+sin θ

( )

Hence,sumoftheseriesupto15termsis,

S15 =15 ×16

15 ⋅16

2+31

3

=60 ×120 +60×31

3

=7200 +620 =7820

{ }

sin22θ +cos42θ =3

4

⇒1−cos22θ +cos42θ =3

4

⇒cos22θ(1−cos22θ) = 1

4...(i)

∵G.M.≤A.M.

∴(cos22θ)(1−cos22θ) ≤ cos22θ + ( 1−cos22θ



4...(ii)

So,fromequation(i)and(ii),weget.

G.M.=A.M.

Itispossibleonlyif

cos22θ =1−cos22θ

⇒cos22θ =1

2⇒cos 2 θ = ± 1

√2

⇒θ=π

8,3π

8∴Sum = π

8+3π

8=π

( )

B. 1

1+cos θ +1

1−sin θ

C. 1

1−cos θ +1

1+sin θ

D. 1

1+cos θ −1

1−sin θ

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question174

Leta1,a2, ..., a10beaG.P.If a3

a1=25,then a9

a5equals:

[Jan.11,2019(I)]

Options:

A.54

B.4(52)

C.53

D.2(52)

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question175

x2sin θ −x(sin θ ⋅cos θ +1) + cos θ =0.

x2sin θ −x sin θ ⋅cos θ −x+cos θ =0

x sin θ(x−cos θ) − 1(x−cos θ) = 0

(x−cos θ)(x sin θ −1) = 0

∴x=cos θ,cosecθ,θ∈ (0,45∘)

∴α=cos θ,β=cosecθ

∞

∑

n=0

αn=1+cos θ +cos2θ+ .. . ∞=1

1−cos θ

∞

∑

n=0

(−1)n

βn=1−1

cosecθ +1

cosec2θ−1

cosec3θ+ .. . ∞

=1−sin θ +sin2θ−sin3θ+ .. . ∞

1+sin θ

∴

∞

∑

n=0

αn+(−1)n

βn=

∞

∑

n=0

αn+

∞

∑

n=0

(−1)n

βn

1−cos θ +1

1+sin θ

( )

Leta1=a,a2=ar,a3=ar2...a10 =ar9

wherer= commonratioofgivenG.P.

Given, a3

=25

⇒ar2

a=25

⇒r= ±5

Now, a9

=ar8

ar4=r4= (±5)4=54

Thesumofaninfinitegeometricserieswithpositivetermsis3andthesum

ofthecubesofitstermsis 27

19.Thenthecommonratioofthisseriesis:

[Jan.11,2019(I)]

Options:

A. 1

B. 2

C. 2

D. 4

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question176

LetSn=1+q+q2+ .... + qnandT n=1+q+1

2+q+1

2+ ... + q+1

whereqisarealnumberandq ≠1.If

101C1+101C2⋅S1+ .... + 101C101 ⋅S100 =αT 100,thenαisequalto:

[Jan.11,2019(II)]

Options:

A.299

B.202

C.200

D.2100

Answer:D

Solution:

( ) ( ) ( )

Letthetermsofinfiniteseriesarea,ar,ar2,ar3, ...

So, a

1−r=3

Since,sumofcubesofitstermsis 27

19thatissumofa3,

a3r3, .. . ∞is 27

19

So, a3

1−r3=27

19

⇒a

1−r×a2

(1+r2+r)=27

19

⇒9(1+r2−2r) × 3

1+r2+r=27

19

⇒6r2−13r +6=0

⇒(3r −2)(2r −3) = 0

⇒r=2

3,or 3

2

As|r| < 1

So,r=2

-------------------------------------------------------------------------------------------------

Question177

TheproductofthreeconsecutivetermsofaG.P.is512.If4isaddedtoeach

ofthefirstandthesecondoftheseterms,thethreetermsnowformanA.P.

ThenthesumoftheoriginalthreetermsofthegivenG.P.is:

[Jan.12,2019(I)]

Options:

A.36

B.32

C.24

D.28

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Sn=1−qn+1

1−q,Tn=

1−q+1

n+1

1−q+1



⇒T100 =

1−q+1

101

1−q+1



Sn =1

1−q−qn+1

1−q,T100 =2101 − (q+1)101

2100(1−q)

Now,101C1+101C2S1+101C3S2+ ... + 101C101S100

1−q(101C2+ ... + 101C101)

−1

1−q(101C2q2+101C3q3+ ... + 101C101q101) + 101

1−q(2101 −1−101) − 1

1−q( (1+q)101 −1

−101C1q) + 101

1−q[2101 −102 − (1+q)101 +1+101q] + 101

1−q[2101 −101 +101q − (1+q)101] + 101

1−q[2101 − (1+q)101] = 2100T100

Hence,bycomparisonα=2100

( ) ( )

( )

LetthreetermsofaG.P.be a

r,a,ar

r⋅a⋅ar =512

a3=512 ⇒a=8

4isaddedtoeachofthefirstandthesecondofthreetermsthenthreetermsare, 8

r+4,8+4,8r.

∵8

r+4,12,8rformanA.P.

∴2×12 =8

r+8r +4

⇒2r2−5r +2=0

⇒(2r −1)(r−2) = 0

⇒r=1

2or2

Therefore,sumofthreeterms = 8

2+8+16 =28

Question178

LetSk=1+2+3+ ..... + k

kIfS1

2+S2

2+ ..... + S10

2=5

12A.ThenAisequalto

[Jan.12,2019(I)]

Options:

A.283

B.301

C.303

D.156

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question179

Ifthesumofthefirst15termsoftheseries

3+11

3+21

3+33+33

3+ ......isequalto225kthenkisequal

to:

[Jan.12,2019(II)]

Options:

A.108

B.27

C.54

D.9

Answer:B

Solution:

( ) ( ) ( ) ( )

∵1+2+3+ ... + k=k(k+1)

2

∴Sk=k(k+1)

2k =k+1

2

⇒5

12A=1

4[22+33+ ... + 112]

4[12+22+ ... + 112−1]

11(11 +1)(2×11 +1)

6−1

11 ×12 ×23

6−1

4[505]

A=505

4×12

5=303

[ ]

S=3

3+3

3+9

3+ (3)3+ ...

S=3

3+6

3+9

3+12

3+ ...

LetthegeneraltermofSbe

( ) ( ) ( )

( ) ( ) ( ) ( )

-------------------------------------------------------------------------------------------------

Question180

I f nC4,nC5andnC6areinA.P.,thenncanbe

[Jan.12,2019(II)]

Options:

A.9

B.14

C.11

D.12

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question181

If19thtermofanon-zeroA.P.iszero,thenits(49thterm):(29thterm)is:

[Jan.11,2019(II)]

Options:

A.4:1

B.1:3

C.3:1

D.2:1

Answer:C

Solution:

Tr=3r

3,then

255K =

∑

r=1

Tr=3

315

∑

r=1

r3

255K =27

64 ×15 ×16

2

⇒K=27

( )

SincenC4,nC5andnC6areinA.P.

2nC5=nC4+nC6

nC4

nC5

nC6

nC5



2=5

n−4+n−5

5

⇒12(n−4) = 30 +n2−9n +20

⇒n2−21n +98 =0

(n−7)(n−14) = 0

n=7,n=14

LetfirsttermandcommondifferenceofAPbeaanddrespectively,then

tn=a+ (n−1)d

∴t19 =a+18d =0∴ a= −18d 

-------------------------------------------------------------------------------------------------

Question182

Thesumofalltwodigitpositivenumberswhichwhendividedby7yield2or

5asremainderis:

[Jan.10,2019(I)]

Options:

A.1256

B.1465

C.1365

D.1356

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question183

Leta1,a2, ......, a30beanA.P.,S =30

∑

i=1aiandT =15

∑

i=1a(2i −1)Ifa5=27and

S−2T =75,thena10isequalto:

[Jan.09,2019(I)]

Options:

A.52

B.57

C.47

D.42

Answer:A

Solution:

∴t49

t29

=a+48d

a+28d 

=−18d +48d

−18d +28d =30d

10d =3

t49 :t29 =3:1

Twodigitpositivenumberswhichwhendividedby7yield2asremainderare12termsi.e,16,23,30, ..., 93

Twodigitpositivenumberswhichwhendividedby7yield

5asremainderare13termsi.e,12,19,26, ..., 96

ByusingAPsumof16,23, ..., 93,weget

S1=16 +23 +30 + ... + 93 =654

ByusingAPsumof12,19,26, ..., 96,weget

S1=12 +19 +26 + ... + 96 =702

∴requiredSum =S1+S2=654 +702 =1356

∑

i=1

ai=30

2[2a1+29d ]

∑

i=1

a(2i −1)=15

2[2a1+28d ]

Since,S−2T =75

-------------------------------------------------------------------------------------------------

Question184

Ifthreedistinctnumbersa,b,careinG.P.andtheequations

ax2+2bx +c=0andd x2+2ex +f=0haveacommonroot,thenwhichone

ofthefollowingstatementsiscorrect?

[April08,2019(II)]

Options:

A. d

a,e

b,f

careinA.P.

B.d ,e,f areinA.P.

C.d ,e,f areinG.P.

D. d

a,e

b,f

careinG.P.

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question185

Forx ∈R,let[x]denotethegreatestinteger≤x,thenthesumoftheseries

−1

3+ − 1

3−1

100 + − 1

3−2

100 + ⋅s+ − 1

3−99

100 is

[April12,2019(I)]

Options:

A.-153

B.-133

C.-131

D.-135

Answer:B

[ ] [ ] [ ] [ ]

⇒30a1+435d −30a1−420d =75

⇒d=5

Also,a5=27 ⇒a1+4d =27 ⇒a1=7

Hence,a10 =a1+9d =7+9×5=52

Sincea,b,careinG.P.

∴b2=acc

Givenequationis,ax2+2bx +c=0

⇒ax2+2√acx +c=0⇒ (√ax + √c)2=0

⇒x= − c

a

Also,giventhatax2+2bx +c=0andd x2+2ex +f=0haveacommonroot.

⇒x= − c

amustsatisfyd x2+2ex +f=0

⇒d⋅c

a+2e −c

a+f=0

a−2e

√ac +f

c=0⇒d

a−2e

b+f

c=0

⇒2e

b=d

a+f

c⇒d

a,e

b,f

careinA.P.

√

(√)

Solution:

-------------------------------------------------------------------------------------------------

Question186

Thesum 3×13

12+5× (13+23)

12+22+7× (13+23+33)

12+22+32+ ......upto10 th term,is:

[April10,2019(I)]

Options:

A.680

B.600

C.660

D.620

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question187

Thesum1 +13+23

1+2+13+23+33

1+2+3+ ......+13+23+33+ ... + 153

1+2+3+ ... + 15 −1

2(1+2+3+ ... + 15is

equalto:

[April10,2019(II)]

Options:

A.620

B.1240

C.1860

D.660

)

∵[x] + x+1

n+x+2

n... . x+n−1

n= [nx]

and[x] + [−x] = −1(x∉z)

∴ − 1

3+ − 1

3−100 + ... + − 1

3−99

100 

= −100 −1

3+1

100 + .. . 1

3+99

100 

= −100 −100

3= −133

[ ] [ ] [ ]

{ [ ] [ ] [ ] }

[ ]

rth termoftheseries,

Tr=(2r +1)(13+23+33+ ... + r3)

12+22+32+ ... + r2

Tr= (2r +1)r(r+1)

2×6

r(r+1)(2r +1)=3r(r+1)

2

∴sumof10termsis = S=

∑

r=1

Tr=3

∑

r=1

(r2+r)

10 × (10 +1)(2×10 +1)

6+10 ×11

2

2×5×11 ×8=660

( )

{ }

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question188

Thesumoftheseries1 +2×3+3×5+4×7+ ......upto11 th termis:

[April09,2019(II)]

Options:

A.915

B.946

C.945

D.916

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question189

Someidenticalballsarearrangedinrowstoformanequilateraltriangle.The

firstrowconsistsofoneball,thesecondrowconsistsoftwoballsandsoon.

If99moreidenticalballsareaddedtothetotalnumberofballsusedin

formingtheequilateraltriangle,thenalltheseballscanbearrangedina

squarewhoseeachsidecontainsexactly2ballslessthanthenumberofballs

eachsideofthetrianglecontains.Thenthenumberofballsusedtoformthe

equilateraltriangleis:

[April09,2019(II)]

Options:

A.157

Let,S=1+13+23

1+2+13+23+33

1+2+3+ .. . 15terms

Tn=13+23+ .. . n3

1+2+ .. . n=

n(n+1)

=n(n+1)

2

Now,S=1

∑

n=1

n2+

∑

n=1

15(16)(31)

6+15(16)

2

=680

∴requiredsumis,680 −1

15(16)

2=680 −60 =620

( )

( ) ( )

1+2.3 +3.5 +4.7 + .......Let,S= (2.3 +3.5 +4.7 + ......)

Now,S10 =

∑

n=1

(n+1)(2n +1) =

∑

n=1

(2n2+3n +1)

=2n(n+1)(2n +1)

6+3n(n+1)

2+n

Putn=10

=2.10.11.21

6+3.10.11

2+10 =945

Hencerequiredsumoftheseries = 1+945 =946

B.262

C.225

D.190

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question190

Thesum 20

∑

k=1k1

2kisequalto:

[April08,2019(II)]

Options:

A.2 −3

217

B.1 −11

220

C.2 −11

219

D.2 −21

220

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Numberofballsusedinequilateraltriangle

=n(n+1)

2

∵sideofequilateraltrianglehasn-balls

∴no.ofballsineachsideofsquareis = (n−2)

Accordingtothequestion, n(n+1)

2+99 = (n−2)2

⇒n2+n+198 =2n2−8n +8

⇒n2−9n −190 =0⇒ (n−19)(n+10) = 0

⇒n=19

Numberofballsusedtoformtriangle

=n(n+1)

2=19 ×20

2=190

Let,S=

∑

k=1

k⋅1

2k

S=1

2+2⋅1

22+3⋅1

23+ .... + 20⋅1

220...(i)

2S=1

22+2⋅1

23+ .... + 19 1

220 +20 1

221...(ii)

Onsubtractingequations(ii)by(i),

2=1

2+1

22+1

23+ .... + 1

220 −20 1

221

21−1

220

1−1

−20⋅1

221 =1−1

220 −10⋅1

220

2=1−11⋅1

220 ⇒S=2−11⋅1

219 =2−11

219

( )

Question191

LetSndenotethesumofthefirstntermsofanA.P.IfS4=16andS6= −48,

thenS10isequalto:

[April12,2019(I)]

Options:

A.-260

B.-410

C.-320

D.-380

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question192

Ifa1,a2,a3, ......areinA.P.suchthata1+a7+a16 =40,thenthesumofthe

first15termsofthisA.P.is:

[April12,2019(II)]

Options:

A.200

B.280

C.120

D.150

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Given,S4=16andS6= −48

⇒2(2a +3d ) = 16 ⇒2a +3d =8...(i)

And3[2a +5d ] = −48 ⇒2a +5d = −16

⇒2d = −24 [usingequation(i)]

⇒d= −12anda=22

∴S10

2= (44 +9(−12)) = −320

LetthecommondifferenceoftheA.P.is'd'.

Given,a1+a7+a16 =40

⇒a1+a1+6d +a1+15d =40

⇒3a1+21d =40

⇒a1+7d =40

3

Now,sumoffirst15termsofthisA.P.is,

S15 =15

2[2a1+14d ] = 15(a1+7d )

=15 40

3=200

[Using(i)]

( )

Question193

Ifa1,a2,a3, .... . anareinA.P.anda1+a4+a7+ .... + a16 =114then

a1+a6+a11 +a16isequalto:

[April10,2019(I)]

Options:

A.98

B.76

C.38

D.64

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question194

Letthesumofthefirstntermsofanon-constantA.P.

a1,a2,a3, ..............be50n +n(n−7)

2A,whereAisaconstant.Ifd isthe

commondifferenceofthisA.P.,thentheorderedpair(d,a50)isequalto:

[April09,2019(I)]

Options:

A.(50,50 +46A)

B.(50,50 +45A)

C.(A,50 +45A)

D.(A,50 +46A)

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question195

Let ∑

k=1

10 f(a+k) = 16(210 −1),wherethefunctionf satisfies

a1+a4+a7+ ... + a16 =114

⇒3(a1+a16) = 114

⇒a1+a16 =38

Now,a1+a6+a11 +a16 =2(a1+a16) = 2×38 =76

∵Sn=50 −7A

2n+n2×A

2⇒a1=50 −3S

∴d=a2−a1=Sn2

−Sn1

⇒d=A

2×2=A

Now,a50 =a1+49 ×d

= (50 −3A) + 49A =50 +46A

So,(d,a50) = (A,50 +46A)

( )

f(x+y) = f(x)f(y)forallnaturalnumbersx,yandf (a) = 2Thenthenatural

number'a'is:

[April09,2019(I)]

Options:

A.2

B.16

C.4

D.3

Answer:D

Solution:

Question196

IfthesumandproductofthefirstthreetermsinanA.P.are33and1155,

respectively,thenavalueofits11 th termis:

[April09,2019(II)]

Options:

A.-35

B.25

C.-36

D.-25

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

∵f(x+y) = f(x) ⋅ f(y)

⇒Letf (x) = tx

∵f(1) = 2

∴t′=2

⇒f(x) = 2x

Since,

∑

k=1

f(a+k) = 16(210 −1)

Then,

∑

k=1

2a+k=16(210 −1)

⇒2a10

∑

k=1

2k=16(210 −1)

⇒2a×((210) − 1) × 2

(2−1)=16 × (210 −1)2.2a=16

⇒a=3

LetthreetermsofA.P.area−d,a,a+d

Sumoftermsis,a−d+a+a+d=33 ⇒a=11

Productoftermsis,(a−d)a(a+d) = 11(121 −d2) = 1155

⇒121 −d2=105 ⇒d= ±4ifd=4

T11 =T1+10d =7+10(4) = 47

ifd= −4

T11 =T1+10d =15 +10(−4) = −25

Question197

Thesumofallnaturalnumbers'n'suchthat100 <n<200andH.C.F.

(91,n) > 1is:

[April08,2019(I)]

Options:

A.3203

B.3303

C.3221

D.3121

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question198

Ifα,βandγarethreeconsecutivetermsofanonconstantG.P.suchthatthe

equationsαx2+2βx +γ=0andx2+x−1=0haveacommonroot,then

α(β+γ)isequalto:

[April12,2019(II)]

Options:

A.0

B.αβ

C.αγ

D.βγ

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

∵91 =13 ×7

Then,therequirednumbersareeitherdivisibleby7or13.

∴Sumofsuchnumbers=Sumofno.divisibleby7+

sumoftheno.divisibleby13−Sumofthenumbersdivisibleby91

= (105 +112 + ... + 196) + (104 +117 + ... + 195) − 182

=2107 +1196 −182 =3121

∵α,β,γarethreeconsecutivetermsofanon-constantG.P.

∴β2=αγ

Sorootsoftheequationαx2+2βx +γ=0are

−2β ±2β2−αγ

2α =β

α

∵αx2+2βx +γ=0andx2+x−1=0haveacommonroot.

∴thisrootsatisfytheequationx2+x−1=0

β2−αβ −α2=0

⇒αγ −αβ −α2=0⇒α+β=γ

Now,α(β+γ) = αβ +αγ

=αβ +β2= (α+β)β=βγ

√

Question199

Leta,bandcbeinG.P.withcommonratior,wherea ≠0and0 <r≤1

2.If

3a,7band15carethefirstthreetermsofanA.P.,thenthe4 th termofthis

A.P.is:

[April10,2019(II)]

Options:

A. 2

B.5a

C. 7

D.a

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question200

Let 1

, ...., ( xi≠0fori =1,2, ..., n)beinA.P.

suchthatx1=4andx21 =20.Ifnistheleastpositiveintegerforwhich

xn>50,then n

∑

i=1

isequalto.

[OnlineApril16,2018]

Options:

A.3

B. 13

C. 13

D. 1

Answer:C

Solution:

( )

∵a,b,careinG.P.⇒b=ar,c=ar2

∵3a,7b,15careinA.P.

⇒3a,7ar,15ar2areinA.P.

∴14ar =3a +15ar2

⇒15r2−14r +3=0⇒r=1

3or 3

5

∵r<1

2∴r=3

5rejected

Fourthterm = 15ar2+7ar −3a

=a(15r2+7r −3) = a15

9+7

3−3=a

( )

-------------------------------------------------------------------------------------------------

Question201

Ifx1,x2, ...., xnand 1

h2, ...⋅ 1

aretwoA.P'ssuchthatx3=h2=8and

x8=h7=20,thenx5.h10equals.

[OnlineApril15,2018]

Options:

A.2560

B.2650

C.3200

D.1600

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question202

∵1

, ...., 1

areinA.P

x1=4andx21 =20

Let′d'bethecommondifferenceofthisA.P

its21 st term = 1

x21

+ [(21 −1) × d]

⇒d=1

20 ×1

20 −1

4⇒d= − 1

100

Also xn>50(given).

∴1

+ [(n−1) × d]

⇒xn=x1

1+ (n−1) × d×x1



∴x1

1+ (n−1) × d×x1

>50

⇒4

1+ (n−1) × − 1

100 ×4

>50

⇒1+ (n−1) × − 1

100 ×4<4

50

⇒− 1

100(n−1) < − 23

100

⇒n−1>23 ⇒n>24

Therefore,n=25.

⇒

∑

i=1

=25

22×1

4+ (25 −1) × − 1

100 =13

( )

[ ( ) ( ) ]

Supposed1isthecommondifferenceoftheA.P.x1,x2, ... . xnthen

∵x8−x3=5d 1=12 ⇒d1=12

5=2.4

⇒x5=x3+2d 1=8+2×12

5=12.8

Supposed2isthecommondifferenceoftheA.P

, ...⋅ 1

then

5d 2=1

20 −1

8=−3

40 ⇒d2=−3

200

∵1

h10

+3d 2=1

200 ⇒h10 =200

⇒x5⋅h10 =12.8 ×200 =2560

IfbisthefirsttermofaninfiniteG.Pwhosesumisfive,thenbliesinthe

interval.

[OnlineApril15,2018]

Options:

A.(−∞, −10)

B.(10,∞)

C.(0,10)

D.(-10,0)

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question203

LetAn=3

4−3

2+3

3− ... + (−1)n−13

nandBn=1−An.Then,the

leastoddnaturalnumberp,sothatBn>An,foralln ≥pis

[OnlineApril15,2018]

Options:

A.5

B.7

C.11

D.9

Answer:B

Solution:

( ) ( ) ( ) ( )

Firstterm = bandcommonratio = r

Forinfiniteseries,Sum =b

1−r=5

⇒b=5(1−r)

So,intervalofb= (0,10)as,−1<r<1forinfiniteG.P.

An=3

4−3

2+3

3− ... + (−1)n−13

n

WhichisaG.P.witha=3

4,r=−3

4andnumberofterms = n

∴An=

4×1−−3

1−−3

4×1−−3



⇒An=3

71−−3

n...(i)

As,Bn=1−An

Forleastoddnaturalnumberp,suchthatBn>An

⇒1−An>An⇒1>2×An⇒An<1

2

Fromeqn.(i),weget

7×1−−3

n<1

2⇒1−−3

n<7

6

⇒1−7

6<−3

n⇒−1

6<−3

n

( ) ( ) ( ) ( )

( ( ) )

( )

( ( ) )

[ ( ) ]

[ ( ) ] ( )

( ) ( )

-------------------------------------------------------------------------------------------------

Question204

Ifa,b,careinA.P.anda2,b2,c2areinG.P.suchthata <b<cand

a+b+c=3

4,thenthevalueofais

[OnlineApril15,2018]

Options:

A. 1

4−1

3√2

B. 1

4−1

4√2

C. 1

4−1

√2

D. 1

4−1

2√2

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question205

Thesumofthefirst20termsoftheseries1 +3

2+7

4+15

8+31

16 + ...is?

[OnlineApril16,2018]

Options:

A.38 +1

220

Asnisodd,then −3

n= − 3n

4

So −1

6< − 3

n⇒1

6>3

n

log 1

6=n log 3

4⇒6.228 <n

Hence,nshouldbe7.

( )

( ) ( )

∵a,b,careinA.P.thena+c=2b

alsoitisgiventhat,

a+b+c=3

4...(i)

⇒2b +b=3

4⇒b=1

4...(ii)

Againitisgiventhat,a2,b2,c2areinG.P.then

(b2)2=a2c2⇒ac = ± 1

16...(iii)

From(i),(ii)and(iii),weget;

a±1

16a =1

2⇒16a2−8a ±1=0

CaseI:16a2−8a +1=0

⇒a=1

4notpossibleasa<b)

CaseII:16a2−8a −1=0⇒a=8± √128

32 

⇒a=1

4±1

2√2

∴a=1

4−1

2√2(∵a<b)

(

B.39 +1

219

C.39 +1

220

D.38 +1

219

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question206

LetAbethesumofthefirst20termsandBbethesumofthefirst40terms

oftheseries12+2⋅22+32+2⋅42+52+2⋅62+ ...IfB −2A =100λ,thenλis

equalto

[2018]

Options:

A.248

B.464

C.496

D.232

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question207

Thegeneraltermofthegivenseries = 2×2r−1

2r,wherer≥0

∴req.sum =1+

∑

r=1

2×2r−1

2r

Now,

∑

r=1

2×2r−1

2r=

∑

r=1

2−1

2r

=2(19) −

21−1

1−1

=38 +

19 −1

1

=38 +1

19 −1=37 +1

19

∴req.sum =1+37 +1

19 =38 +1

( ) ( )

( ( ) ) ( )

( ) ( )

Here,B−2A

∑

n=1

an−2

∑

n=1

an=

∑

n=21

an−2

∑

n=1

an

B−2A = (212+2.222+232+2.242+ ..... + 402)

−(12+2.22+32+2.42..... + 202)

=20[22 +2.24 +26 +2.28 + ..... + 60]

=20[(22 +24 +26...60)20 terms + (24 +28 + .... + 60)10 terms ]

20 20

2(22 +60) + 10

2(24 +60) 

=10[20.82 +10.84]

=100[164 +84] = 100.248

[ ]

Leta1,a2,a3, ..., a49beinA.P.suchthat 12

∑

k=0a4k +1=416anda9+a43 =66.If

2+a2

2+ ... + a17

2=140m,thenmisequalto

[2018]

Options:

A.68

B.34

C.33

D.66

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question208

Foranythreepositiverealnumbersa,bandc,

9(25a2+b2) + 25(c2−3ac) = 15b(3a +c).Then

[2017]

Options:

A.a,bandcareinG.P.

B.b,candaareinG.P.

C.b,candaareinA.P.

D.a,bandcareinA.P.

Answer:C

Solution:

∵∑

k=0

12 a4k +1=416 ⇒13

2[2a1+48d ] = 416

⇒a1+24d =32

N ow,a9+a43 =66 ⇒2a1+50d =66

Fromeq.(i)&(ii)weget;d=1anda1=8

Also,

∑

r=1

∑

r=1

[8+ (r−1)1]2=140m

⇒

∑

r=1

(r+7)2=140m

⇒

∑

r=1

(r2+14r +49) = 140m

⇒17 ×18 ×35

6+14 17 ×18

2+ (49 ×17) = 140

⇒m=34

( ) ( )

\Wehave

9(25a2+b2) + 25(c2−3ac) = 15b(3a +c)

⇒225a2+9b2+25c2−75ac =45ab +15bc

⇒(15a)2+ (3b)2+ (5c)2−75ac −45ab −15bc =0

2[(15a −3b)2+ (3b −5c)2+ (5c −15a)2] = 0

Itispossiblewhen15a −3b =0,3b −5c =0and5c −15a =0

⇒15a =3b ⇒b=5a

-------------------------------------------------------------------------------------------------

Question209

Ifthreepositivenumbersa,bandcareinA.P.suchthatabc =8,thenthe

minimumpossiblevalueofbis:

[OnlineApril9,2017]

Options:

A.2

B.4

C.4

D.4

Answer:A

Solution:

Question210

Ifthearithmeticmeanoftwonumbersaandb,a>b>0,isfivetimestheir

geometricmean,then a+b

a−bisequalto:

[OnlineApril8,2017]

Options:

A. √6

B. 3√2

C. 7√3

D. 5√6

Answer:D

Solution:

⇒b=5c

3,a=c

3

⇒a+b=c

3+5c

3=6c

3

⇒a+b=2c

⇒b,c,aareinA.P.

A.T.Q.

A.M.=5G.M.

ByArithmeticMean:

a+c=2b

Considera=b=c=2

⇒abc =8

⇒a+b=2b

∴minimumpossiblevalueofb=2

-------------------------------------------------------------------------------------------------

Question211

Leta,b,c∈R.Iff(x) = ax2+bx +cissuchthata +b+c=3and

f(x+y) = f(x) + f(y) + xy, ∀x,y∈R,then ∑

n=1

10 f(n)isequalto:

[2017]

Options:

A.255

B.330

C.165

D.190

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question212

LetSn=1

13+1+2

13+23+1+2+3

13+23+33+ ....+1+2+ ...... + n

13+23+ ...... + n3,If100Sn=n,thennisequal

to:

[OnlineApril9,2017]

Options:

A.199

a+b

2=5√ab

a+b

√ab =10

∴a

b=10 + √96

10 − √96 =10 +4√6

10 −4√6

UseComponendoandDividendo

a+b

a−b=20

8√6=5

2√6=5√6

f(x) = ax2+bx +c

f(1) = a+b+c=3⇒f(1) = 3

Now f (x+y) = f(x) + f(y) + xy ...

Putx=y=1ineqn(i)

f(2) = f(1) + f(1) + 1=2f (1) + 1

f(2) = 7

⇒f(3) = 12

Sn=3+7+12 + ....... + tn

Sn=3+7+12 + ....... + tn−1+tn

0=3+4+5.........tonterm−tn

tn=3+4+5+ ...uptonterms

tn=(n2+5n)

2

Sn= ∑ tn= ∑ (n2+5n)

2

Sn=1

n(n+1)(2n +1)

6+5n(n+1)

2

=n(n+1)(n+8)

6

S10 =10 ×11 ×18

6=330

[ ]

B.99

C.200

D.19

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question213

Ifthesumofthefirstntermsoftheseries√3+ √75 + √243 + √507 + .....is

435√3,thennequals:

[OnlineApril8,2017]

Options:

A.18

B.15

C.13

D.29

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question214

Leta1,a2,a3, ..., an,beinA.P.Ifa3+a7+a11 +a15 =72,thenthesumofits

Tn=

n(n+1)

2

⇒Tn=2

n(n+1)

⇒Sn= ∑ Tn=2

∑

n=1

n−1

n+1=2 1 −1

n+1

⇒Sn=2n

n+1

∵100Sn=n

⇒100×2n

n+1=n

⇒n+1=200

⇒n=199

( )

( ) { }

∵√3[1+ √25 + √81 + √69 + .......] = 435√3

⇒√3[1+5+9+13 + ..... + Tn] = 435√3

⇒√3×n

2[2+ (n−1)4] = 435√3

⇒2n +4n2−4n =870

⇒4n2−2n −870 =0

⇒2n2−n−435 =0

n=1± √1+4×2×435

4=1±59

4

∴n=1+59

4=15;orn=1−59

4=14.5

first17termsisequalto:

[OnlineApril10,2016]

Options:

A.306

B.204

C.153

D.612

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question215

Ifthe2 nd ,5th and9 th termsofanon-constantA.P.areinG.P.,thenthe

commonratioofthisG.P.is:

[2016]

Options:

A.1

B. 7

C. 8

D. 4

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question216

Letz =1+aibeacomplexnumber,a >0,suchthatz3isarealnumber.Then

thesum1 +z+z2+ .... + z11isequalto:

(OnlineApril10,2016)

a3+a7+a11 +a15 =72

(a3+a15) + (a7+a11) = 72

a3+a15 +a7+a11 =2(a1+a17)

a1+a17 =36

S17 =17

2[a1+a17] = 17 ×18 =306

LettheGPbea,arandar2thena=A+d;ar =A+4d ;

ar2=A+8d 

⇒ar2−ar

ar −a=(A+8d ) − (A+4d )

(A+4d ) − (A+d)

r=4

Options:

A.1365√3i

B.−1365√3i

C.−1250√3i

D.1250√3i

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question217

IfA >0,B>0andA +B=π

6,thentheminimumvalueoftan A +tan Bis:

[OnlineApril10,2016]

Options:

A.√3− √2

B.4 −2√3

C. 2

√3

D.2 − √3

Answer:B

Solution:

z=1+ai

z2=1−a2+2ai

z2⋅z= {(1−a2) + 2ai} {1+ai}

= (1−a2) + 2ai + (1−a2)ai −2a2

∵z3isreal⇒2a + (1−a2)a=0

a(3−a2) = 0⇒a= √3(a>0)

1+z+z2......... . z11 =z12 −1

z−1=(1+ √3i)12 −1

1+ √3i−1

=(1+ √3i)12 −1

√3i

(1+ √3i)12 =212 1

2+√3



=212 cos π

3+i sin π

12 =212(cos 4 π +i sin 4 π) = 212

⇒212 −1

√3i=4095

√3i= − 4095

3√3i= −1365√3i

( )

tan(A+B) = tan A +tan B

1−tan A tan B

⇒1

√3=y

1−tan A tan Bwherey=tan A +tan B

⇒tan A tan B =1− √3y

AlsoAM ≥GM 

⇒tan A +tan B

2≥ √tan A tan B

⇒y≥2 1 − √3y

⇒y2≥4−4√3y

⇒y2+4√3y −4≥0

⇒y≤ −2√3−4ory≥ −2√3+4

√

-------------------------------------------------------------------------------------------------

Question218

Letx,y,zbepositiverealnumberssuchthatx +y+z=12and

x3y4z5= (0.1)(600)3.Thenx3+y3+z3isequalto:

[OnlineApril9,2016]

Options:

A.342

B.216

C.258

D.270

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question219

Ifthesumofthefirsttentermsoftheseries

2+22

2+31

2+42+44

2+ ......,is 16

5mthenmisequalto:

[2016]

Options:

A.100

B.99

C.102

D.101

Answer:D

Solution:

( ) ( ) ( ) ( )

(y≤ −2√3−4isnotpossibleastan A tan B >0)

x+y+z=12

AM ≥GM 

3+4y

4+5z

12 ≥12 x

5

x3y4z5

334455≤1

x3y4z5≤33⋅44⋅55

x3y4z5≤ (0.1)(600)3

But,givenx3y4z5= (0.1)(600)3

∴allthenumberareequal

∴x

3=y

4=z

5( = k)

x=3k;y=4k;z=5k

x+y+z=12

3k +4k +5k =12

k=1∴ x=3;y=4;z=5

∴x3+y3+z3=216

( ) ( ) ( ) √( ) ( ) ( )

-------------------------------------------------------------------------------------------------

Question220

Forx ∈R,x≠ −1,if(1+x)2016 +x(1+x)2015 +x2

(1+x)2014 + .... + x2016 =∑

i=0

2016 aixi,thena17isequalto:

[OnlineApril9,2016]

Options:

A. 2017!

17!2000!

B. 2016!

17!1999!

C. 2016!

16!

D. 2017!

2000!

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question221

IfmistheA.M.oftwodistinctrealnumbersl andn(l,n>1)andG1,G2and

G3arethreegeometricmeansbetweenl andn,thenG1

4+2G2

4+G3

4equals.

[2015]

Options:

A.4l mn2

B.4l 2m2n2

C.4l 2mn

2+12

2+16

2+20

2... + 44

2

S=16

25(22+32+42+ ... + 112)

=16

11(11 +1)(22 +1)

6−1=16

25 ×505 =16

5×101

⇒16

5m=16

5×101

⇒m=101

( ) ( ) ( ) ( ) ( )

( )

S= (1+x)2016 +x(1+x)2015 +x2(1+x)2014

+.... + x2015(1+x) + x2016... (i) 

1+xS=x(1+x)2015 +x2(1+x)2014 + .....+

x2016 +x2017

1+x... (ii) 

Subtracting(i)from(ii)

1+x= (1+x)2016 −x2017

1+x

∴S= (1+x)2017 −x2017

a17 = coefficientofx17 =2017C17 =2017!

17!2000!

( )

D.4l m2n

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question222

Thesumofthe3 rd andthe4 th termsofaG.P.is60andtheproductofits

firstthreetermsis1000.IfthefirsttermofthisG.P.ispositive,thenits7 th

termis:

[OnlineApril11,2015]

Options:

A.7290

B.640

C.2430

D.320

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question223

Thesumoffirst9termsoftheseries.

1+13+23

1+3+13+23+33

1+3+5+ ...

[2015]

Options:

A.142

B.192

C.71

m=l+n

2andcommonratioofG.P.

=r=n

4

∴G1=l3∕4n1∕4,G2=l1∕2n1∕2,G3=l1∕4n3∕4

4+2G2

4+G3

4=l3n+2l 2n2+l n3

=ln(l+n)2=ln × (2m)2=4l m2n

( )

Leta,arandar2bethefirstthreetermsofG.PAccordingtothequestion

a(ar)(ar2) = 1000 ⇒ (ar)3=1000 ⇒ar =10

andar2+ar3=60 ⇒ar(r+r2) = 60

⇒r2+r−6=0

⇒r=2, −3

a=5,a= − 10

3reject)

Hence,T7=ar6=5(2)6=5×64 =320

(

D.96

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question224

If 5

∑

n=1

n(n+1)(n+2)(n+3)=k

3,thenkisequalto

[OnlineApril11,2015]

Options:

A. 1

B. 17

105

C. 55

336

D. 19

112

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question225

Thevalueof ∑

r=16

30 (r+2)(r−3)isequalto:

[OnlineApril10,2015]

Options:

A.7770

B.7785

nth termofseries =

n(n+1)

n2=1

4(n+1)2

Sumofnterm = Σ1

4(n+1)2=1

4[Σn2+2Σn +n]

n(n+1)(2n +1)

6+2n(n+1)

2+n

Sumof9terms

9×10 ×19

6+18 ×10

2+9=384

4=96

[ ]

Generaltermofgivenexpressioncanbewrittenas

Tr=1

n(n+1)(n+2)−1

(n+1)(n+2)(n+3)

ontakingsummationboththeside,weget

∑

r=1

5Tr=1

6−1

6.7.8 =k

3

⇒1

3×1

61−1

56 =k

3⇒1

3×1

6×55

56 =k

3

⇒k=55

336

[ ]

( )

C.7775

D.7780

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question226

Letαandβbetherootsofequationpx2+qx +r=0,p≠0Ifp,q,rarein

A.Pand 1

α+1

β=4,thenthevalueof|α−β|is:

[2014]

Options:

A. √34

B. 2√13

C. √61

D. 2√17

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question227

Thesumofthefirst20termscommonbetweentheseries3

+7+11 +15 + .........and1 +6+11 +16 + ......,is

∑

r=1

(r2−r−6) = 7780

Letp,q,rareinAP

⇒2q =p+r...(i)

Given 1

α+1

β=4

⇒α+β

αβ =4

Wehaveα+β= −q∕pandαβ =r

p

−xq

⇒p

=4⇒q= −4r...(ii)

From(i),wehave

2(−4r) = p+r

p= −9r...(iii)

Now,|α−β| = (α+β)2−4αβ

=−q

2−4r

p=q2−4pr

|p|

From(ii)and(iii)

=16r2+36r2

|−9r|=2√13

√

√( ) √

√

[OnlineApril11,2014]

Options:

A.4000

B.4020

C.4200

D.4220

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question228

GivenanA.P.whosetermsareallpositiveintegers.Thesumofitsfirstnine

termsisgreaterthan200andlessthan220.Ifthesecondterminitis12,

thenits4 th termis:

[OnlineApril9,2014]

Options:

A.8

B.16

C.20

D.24

Answer:C

Solution:

Givenn=20;S20 = ?

Series(1) → 3,7,x11,15,19,23,27,x31,35,39,43,4751,55,59...

Series(2) → 1,6,x11,16,21,26,x31,36,41,46,x51,5661,66,71

Thecommontermsbetweenboththeseriesare11,31,51,71...

AboveseriesformsanArithmeticprogression(A.P).

Therefore,firstterm(a) = 11and

commondifference(d) = 20

Now,Sn=n

2[2a + (n−1)d]

S20 =20

2[2×11 + (20 −1)20]

S20 =10[22 +19 ×20]

S20 =10 ×402 =4020

∴S20 =4020

LetabethefirsttermanddbethecommondifferenceofgivenA.P.

Secondterm,a+d=12...(i)

Sumoffirstnineterms,

S9=9

2(2a +8d ) = 9(a+4d )

GiventhatS9ismorethan200andlessthan220

⇒200 <S9<220

⇒200 <9(a+4d ) < 220

⇒200 <9(a+d+3d ) < 220

Puttingvalueof(a+d)fromequation(i)

200 <9(12 +3d ) < 220

⇒200 <108 +27d <220

⇒200 −108 <108 +27d −108 <220 −108

-------------------------------------------------------------------------------------------------

Question229

ThreepositivenumbersformanincreasingG.P.Ifthemiddleterminthis

G.P.isdoubled,thenewnumbersareinA.P.thenthecommonratioofthe

G.P.is:

[2014]

Options:

A.2 − √3

B.2 + √3

C.√2+ √3

D.3 + √2

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question230

Theleastpositiveintegernsuchthat1 −2

3−2

32− .... − 2

3n−1<1

100 ,is:

[OnlineApril12,2014]

Options:

A.4

B.5

C.6

D.7

Answer:B

Solution:

⇒92 <27d <112

Possiblevalueofdis4

27 ×4=108

Thus,92 <108 <112

Puttingvalueofdinequation(i)

a+d=12

a=12 −4=8

4th term = a+3d =8+3×4=20

Leta,ar,ar2areinG.P.

Accordingtothequestion

a,2ar,ar2areinA.P.

⇒2×2ar =a+ar2

⇒4r =1+r2⇒r2−4r +1=0

r=4± √16 −4

2=2± √3

Sincer>1

∴r=2− √3 isrejected 

Hence,r=2+ √3

-------------------------------------------------------------------------------------------------

Question231

Inageometricprogression,iftheratioofthesumoffirst5termstothesum

oftheirreciprocalsis49,andthesumofthefirstandthethirdtermis35.

Thenthefirsttermofthisgeometricprogressionis:

[OnlineApril11,2014]

Options:

A.7

B.21

C.28

D.42

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question232

Thecoefficientofx50inthebinomialexpansionof

(1+x)1000 +x(1+x)999 +x2(1+x)998 + ...+x1000is:

1−2

3−2

32...⋅ 2

3n−1<1

100

⇒1−2

3+1

32+1

33+ ... 1

3n−1<1

100

⇒

1−21

3n−1

3−1

100

⇒1−23n−1

2.3n<1

100

⇒1−3n−1

3n<1

100

⇒1−1+1

3n<1

100 ⇒100 <3n

Thus,leastvalueofnis5

[ ]

[ ( ) ]

[ ]

AccordingtoQuestion

⇒S5

′=49 \{here,S5= Sumoffirst5termsandS5= Sumoftheirreciprocals)

⇒

a(r5−1)

(r−1)

a−1(r−5−1)

(r−1−1)

=49

⇒a(r5−1) × (r−1−1)

a−1(r−5−1) × (r−1)=49

or a2(1−r5) × (1−r) × r5

(1−r5) × (1−r) × r=49

⇒a2r4=49 ⇒a2r4=72

⇒ar2=7...(i)

Also,given,S1+S3=35

a+ar2=35...(ii)

Nowsubstitutingthevalueofeq.(i)ineq.(ii)

a+7=35

a=28

[OnlineApril11,2014]

Options:

A. (1000)!

(50)!(950)!

B. (1000)!

(49)!(951)!

C. (1001)!

(51)!(950)!

D. (1001)!

(50)!(951)!

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question233

LetGbethegeometricmeanoftwopositivenumbersaandb,andM bethe

arithmeticmeanof 1

aand 1

b.If 1

M:Gis4 :5,thena :bcanbe:

[OnlineApril12,2014]

Options:

A.1:4

B.1:2

C.2:3

D.3:4

Answer:A

Solution:

Letgivenexpansionbe

S= (1+x)1000 +x(1+x)999 +x2(1+x)998 + ...+... + x1000

Put1+x=t

S=t1000 +xt999 +x2(t)998 + ... + x1000

ThisisaG.Pwithcommonratio x

t

t1000 1−x

1001

1−x



(1+x)1000 1−x

1+x

1001

1−x

1+x



=(1+x)1001[(1+x)1001 −x1001]

(1+x)1001 

= [(1+x)1001 −x1001]

Nowcoeffofx50inaboveexpansionisequaltocoeffofx50in(1+x)1001whichis1001C50

=(1001)!

50!(951)!

[ ( ) ]

G= √ab

a+1

2

-------------------------------------------------------------------------------------------------

Question234

If(10)9+2(11)1(108) + 3(11)2(10)7+ ....+10(11)9=k(10)9,thenkisequalto:

[2014]

Options:

A.100

B.110

C. 121

D. 441

100

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question235

M=a+b

2ab 

Giventhat 1

M:G=4:5

2ab

(a+b)√ab =4

5

⇒a+b

2√ab =5

4

⇒a+b+2√ab

a+b−2√ab =5+4

5−4

{UsingComponendo&Dividendo}

⇒(√a)2+ (√b)2+2√ab

(√a)2+ (√b)2−2√ab =9

1

⇒√b+ √a

√b− √a

1⇒√b+ √a

√b− √a=3

1

⇒√b+ √a+ √b− √a

√b+ √a− √b+ √a=3+1

3−1

{UsingComponendo&Dividendo}

a=4

2=2

a=4

1

b=1

4⇒a:b=1:4

( )

√

Giventhat109+2⋅ (11)(10)8+3(11)2(10)7+ ... + 10(11)9 = k(10)9

Letx=109+2⋅ (11)(10)8+3(11)2(10)7+ ... + 10(11)9...(i)

Multipliedby 11

10onboththesides

10x=11.108+2. (11)2⋅ (10)7+ ... + 9(11)9+1110...(ii)

Subtract(ii)from(i),weget

x 1 −11

10 =109+11(10)8+112× (10)7+ ... + 119−1110

⇒ − x

10 =109

10 −1

−1110

⇒ − x

10 = (1110 −1010) − 1110 = −1010

⇒x=1011 =k⋅109Given 

⇒k=100

( )

[( ) ]

ThenumberoftermsinanA.P.iseven;thesumoftheoddtermsinitis24

andthattheeventermsis30.Ifthelasttermexceedsthefirsttermby10 1

thenthenumberoftermsintheA.P.is:

[OnlineApril19,2014

Options:

A.4

B.8

C.12

D.16

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question236

Ifthesum 3

12+5

12+22+7

12+22+32+ ......+upto20termsisequalto k

21 ,thenkis

equalto:

[OnlineApril9,2014]

Options:

A.120

B.180

C.240

D.60

Answer:A

Solution:

Leta,dand2nbethefirstterm,commondifferenceandtotalnumberoftermsofanA.P.respectivelyi.e.a

+(a+d) + (a+2d ) + ... + (a+ (2n −1)d)

No.ofeventerms = n,No.ofoddterms = n

Sumofoddterms:

So=n

2[2a + (n−1)(2d )] = 24

⇒n[a+ (n−1)d] = 24...(i)

Sumofeventerms:

Se=n

2[2(a+d) + (n−1)2d ] = 30

⇒n[a+d+ (n−1)d] = 30...(ii)

Subtractingequation(i)from(ii),weget

nd =6...(iii)

Also,giventhatlasttermexceedsthefirsttermby 21

2

a+ (2n −1)d=a+21

2

2nd −d=21

2

⇒2×6−21

2=d(∵nd =6)

d=3

2

Puttingvalueofdinequation(3)n=6×2

3=4

Totalno.ofterms = 2n =2×4=8

Solution:

-------------------------------------------------------------------------------------------------

Question237

Leta1,a2,a3, ...beanA.P,suchthat a1+a2+ ... + ap

a1+a2+a3+ ... + aq

=p3

q3;p≠q.Then a6

a21

is

equalto:

[OnlineApril9,2013]

Options:

A. 41

B. 31

121

C. 11

D. 121

1861

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question238

Thesumoftheseries:1 +1

1+2+1

1+2+3+ .......upto10terms,is

[OnlineApril9,2013]

Options:

nth termofgivenseriesis

2n +1

n(n+1)(2n +1)

n(n+1)

Letnth term,an=61

n−1

n+1

Sumof20terms,S20 =a1+a2+a3+ .... + a20

S20 =61

1−1

2+61

2−1

3+61

3−1

4+ ...

+61

18 −1

19 +61

19 −1

20 +61

20 −1

21 

S20 =1−1

2+1

2−1

3+1

3−1

4+ ...

18 −1

19 +1

19 −1

20 +1

20 −1

21 

S20 =6 1 −1

21 =120

21 ...(i)

GiventhatS20 =k

21...(ii)

Oncomparing(i)and(ii),weget

k=120

[ ]

( ) ( ) ( )

[ ( ) ( ) ( )

( ) ( ) ( ) ]

( )

a1+a2+a3+ ...... + ap

a1+a2+a3+ ...... + aq

=p3

q3

⇒a1+a2

1⇒a1+ (a1+d) = 8a1

⇒d=6a1

Now a6

a21

=a1+5d

a1+20d 

=a1+5×6a1

a1+20 ×6a1

=1+30

1+120 =31

121

A. 18

B. 22

C. 20

D. 16

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question239

Ifa1,a2,a3, ..., an, ....areinA.P.suchthata4−a7+a10 =m,thenthesumof

first13termsofthisA.P.,is:

[OnlineApril23,2013]

Options:

A.10m

B.12m

C.13m

D.15m

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question240

GivensumofthefirstntermsofanA.P.is2n +3n2.AnotherA.P.isformed

withthesamefirsttermanddoubleofthecommondifference,thesumofn

termsofthenewA.P.is:

[OnlineApril22,2013]

Options:

Tr=1

1+2+3+ ... + r=2

r(r+1)S10 =2

∑

r=1

r(r+1)=2

∑

r=1

r+1

r(r+1)−r

r(r+1)

∑

r=1

r−1

r+1

=21

1−1

2+1

2−1

3+1

3−1

4+ ... + 1

10 −1

11 

=2 1 −1

11 =2×10

11 =20

[ ]

( )

[ ( ) ( ) ( ) ( ) ]

[ ]

Ifdbethecommondifference,then

m=a4−a7+a10 =a4−a7+a7+3d =a7

S13 =13

2[a1+a13] = 13

2[a1+a7+6d ]

=13

2[2a7] = 13a7=13m

A.n +4n2

B.6n2−n

C.n2+4n

D.3n +2n2

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question241

Givenasequenceof4numbers,firstthreeofwhichareinG.P.andthelast

threeareinA.P.withcommondifferencesix.Iffirstandlasttermsofthis

sequenceareequal,thenthelasttermis:

[OnlineApril25,2013]

Options:

A.16

B.8

C.4

D.2

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question242

Thesumoffirst20termsofthesequence0.7,0.77,0.777, .....is

[2013]

Options:

GivenSn=2n +3n2

Now,firstterm = 2+3=5

secondterm = 2(2) + 3(4) = 16

thirdterm = 2(3) + 3(9) = 33

Now,sumgiveninoption(b)onlyhasthesamefirsttermanddifferencebetween2nd and1sttermisdoublealso.

Leta,b,c,dbefournumbersofthesequence.

Now,accordingtothequestionb2=acandc−b=6anda

−c=6

Also,givena=d

∴b2=ac ⇒b2=aa+b

2(∵2c =a+b)

⇒a2−2b2+ab =0

Now,c−b=6anda−c=6

givesa−b=12

⇒b=a−12

∴a2−2b2+ab =0

⇒a2−2(a−12)2+a(a−12) = 0

⇒a2−2a2−288 +48a +a2−12a =0

⇒36a =288 ⇒a=8

Hence,lasttermisd=a=8.

[ ]

A. 7

81(179 −10−20)

B. 7

9(99 −10−20)

C. 7

81(179 +10−20)

D. 7

9(99 +10−20)

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question243

Thevalueof12+32+52+ .................... + 252is

[OnlineApril25,2013]

Options:

A.2925

B.1469

C.1728

D.1456

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

LetS=7

10 +77

100 +777

103+ ......+upto20terms

=71

10 +11

100 +111

103+ ......+upto20terms]

Multiplyanddivideby9

10 +99

100 +999

1000 + .....upto20terms]

1−1

10 +1−1

102+1−1

103

+...... upto 20 terms



920 −

10 1−1

1−1



179

9+1

20 =7

81[179 + (10)−20]

[

[( ) ( ) ( ) ]

[( ( ) ) ]

[ ( ) ]

Consider12+32+52+ ...... + 252

nth termTn= (2n −1)2,n=1, ..... . 13

Now, Sn=

∑

n=1

Tn=

∑

n=1

(2n −1)2

∑

n=1

4n2+

∑

n=1

1−

∑

n=1

4n =4∑n2+13 −4∑n

=4n(n+1)(2n +1)

6+13 −4n(n+1)

2

Putn=13,weget

Sn=26 ×14 ×9+13 −26 ×14

=3276 +13 −364 =2925

[ ]

Question244

Thesumoftheseries:

(b)2+2(d)2+3(6)2+ ... upto 10 termsis:

[OnlineApril23,2013]

Options:

A.11300

B.11200

C.12100

D.12300

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question245

Thesum 3

12+5

12+22+7

12+22+32+ ....upto11-termsis:

[OnlineApril22,2013]

Options:

A. 7

B. 11

C. 11

D. 60

Answer:C

Solution:

22+2(4)2+3(6)2+ ......upto10terms

=22[13+23+33+ ......upto10terms]

=4⋅10 ×11

2=12100

( )

Givensumis

12 +5

12+22+7

12+22+32+ ....

nthterm = Tn

2n +1

n(n+1)(2n +1)

6=6

n(n+1)

orTn=61

n−1

n+1

∴Sn= ∑ Tn=6∑1

n−6∑1

n+1=6n

n−6

n+1

=6−6

n+1=6n

n+1

So,sumupto11termsmeans

S11 =6×11

11 +1=66

12 =33

6=11

[ ]

-------------------------------------------------------------------------------------------------

Question246

Thesumoftheseries12+2.22+32+2.42+52+2.62+ .... + 2(2m)2is

[OnlineMay7,2012]

Options:

A.m(2m +1)2

B.m2(m+2)

C.m2(2m +1)

D.m(m+2)2

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question247

ThedifferencebetweenthefourthtermandthefirsttermofaGeometrical

Progresssionis52.Ifthesumofitsfirstthreetermsis26,thenthesumof

thefirstsixtermsoftheprogressionis

[OnlineMay7,2012]

Options:

A.63

B.189

C.728

D.364

Answer:C

Solution:

Thesumofthegivenseries12+2.22+32+2.42+52

+2.62+ ........ + 2(2m)2is 2m(2m +1)2

2=m(2m +1)2

Leta,ar,ar2,ar3,ar4,ar5besixtermsofaG.P.where'a'isfirsttermandriscommonratio.

Accordingtogivenconditions,wehave

ar3−a=5⇒a(r3−1) = 52...(i)

anda+ar +ar2=26

⇒a(1+r+r2) = 26...(ii)

Tofind:a(1+r+r2+r3+r4+r5)

Consider

a[1+r+r2+r3+r4+r5]

=a[1+r+r2+r3(1+r+r2)]

=a[1+r+r2][1+r3]...(iii)

Divide(i)by(ii),weget

r3−1

1+r+r2=2

weknowr3−1= (r−1)(1+r+r2)

∴r−1=2⇒r=3and a=2

∴a(1+r+r2+r3+r4+r5) = a(1+r+r2)(1+r3)

-------------------------------------------------------------------------------------------------

Question248

Ifa,b,c,d andparedistinctrealnumberssuchthat

(a2+b2+c2)p2−2p(ab +bc +cd ) + (b2+c2+d2) ≤ 0then

[OnlineMay12,2012]

Options:

A.a,b,c,d areinA.P.

B.ab =cd

C.ac =bd

D.a,b,c,d areinG.P.

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question249

Thesumoftheseries 1

1+ √2+1

√2+ √3+1

√3+ √4+ .upto15termsis

[OnlineMay12,2012]

Options:

A.1

B.2

C.3

D.4

Answer:C

Solution:

=2(1+3+9)(1+27) = 26 ×28 =728

Thegivenrelationcanbewrittenas

(a2p2−2abp +b2) + (b2p2+c2−2bpc)+(c2p2+d2−2pcd ) ≤ 0

or (ap −b)2+ (bp −c)2+ (cp −d)2≤0...(i)

Sincea,b,c,dandpareallreal,theinequality(i)ispossibleonlywheneachoffactoriszero.

i.e.,ap −b=0,bp −c=0andcp −d=0

or p =b

a=c

b=d

c

ora,b,c,dareinG.P.

Givenseriesis 1

1+ √2+1

√2+ √3+1

√3+ √4+ ....

nth term = 1

√n+ √n+1

∴15 th term =1

√15 + √16 

Thus,givenseriesupto15termsis

1+ √2+1

√2+ √3+1

√3+ √4+ ...... + 1

√15 + √16 

Thiscanbere-writtenas

-------------------------------------------------------------------------------------------------

Question250

Supposeθandφ(≠0)aresuchthatsec(θ+φ), sec θandsec(θ−φ)areinA.P.

Ifcos θ =k cos φ

2forsomek,thenkisequalto

[OnlineMay19,2012]

Options:

A.±√2

B.±1

C.±1

√2

D.±2

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question251

Thesumoftheseries1 +4

3+10

9+28

27 + ...uptontermsis

[OnlineMay19,2012]

Options:

A. 7

6n+1

6−2

3.2n−1

B. 5

3n−7

6+1

2.3n−1

C.n +1

2−1

2.3n

D.n −1

3−1

3.2n−1

Answer:C

( )

1− √2

−1+√2− √3

−1+√3− √4

−1+ ..... + √15 − √16

−1

(Byrationalization)

= −1+ √2− √2+ √3− √3+ √4+ ..... − √14 + √15

= −1+ √16 = −1+4=3

Hence,therequiredsum = 3

Since,sec(θ−φ), sec θandsec(θ+φ)areinA.P.,∴2 sec θ =sec(θ−φ) + sec(θ+φ)

⇒2

cos θ =cos(θ+φ) + cos(θ−φ)

cos(θ−φ)cos(θ+φ)

⇒2(cos2θ−sin2φ) = cos θ[2 cos θ cos φ]

⇒cos2θ(1−cos φ) = sin2φ=1−cos2φ

⇒cos2θ=1+cos φ =2cos2φ

2

∴cos θ = √2 cos φ

2

Butgivencos θ =k cos φ

2

∴k= √2

Solution:

-------------------------------------------------------------------------------------------------

Question252

IftheA.M.betweenp th andq th termsofanA.P.isequaltotheA.M.

betweenr th ands th termsofthesameA.P.thenp +qisequalto

[OnlineMay26,2012]

Options:

A.r +s−1

B.r +s−2

C.r +s+1

D.r +s

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question253

If100timesthe100 th termofanAPwithnonzerocommondifference

equalsthe50timesits50 th term,thenthe150 th termofthisAPis:

[2012]

Options:

A.-150

B.150timesits50 th term

C.150

D.Zero

Answer:D

Givenseriesis1+4

3+10

9+28

27 + ... . nterms

=1+1+1

3+1+1

9+1+1

27 + ... . nterms

= ( 1+1+1+ .... + nterms)

3+1

9+1

27 + ... . n terms 

=n+

31−1

1−1

=n+1

3×3

2[1−3−n]

=n+1

2[1−3−n] = n+1

2−1

2.3n

( ) ( ) ( )

( )

Given: ap+aq

2=ar+aS

2

⇒a+ (p−1)d+a+ (q−1)d

=a+ (r−1)d+a+ (s−1)d

⇒2a + (p+q)d−2d =2a + (r+s)d−2d 

⇒(p+q)d= (r+s)d⇒p+q=r+s

Solution:

-------------------------------------------------------------------------------------------------

Question254

Statement-1:Thesumoftheseries1 + (1+2+4)+

(4+6+9) + (9+12 +16) + .... + (361 +380 +400)is8000

Statement-2: n

∑

k=1(k3− (k−1)3) = n3,foranynaturalnumbern.

[2012]

Options:

A.Statement-1isfalse,Statement-2istrue.

B.Statement-1istrue,statement-2istrue;statement-2isacorrectexplanationforStatement-1.

C.Statement-1istrue,statement-2istrue;statement-2isnotacorrectexplanationforStatement-1

.

D.Statement-1istrue,statement-2isfalse.

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question255

Letanbethen th termofanA.P.If 100

∑

r=1a2r =αand 100

∑

r=1a2r −1=β,thenthe

commondifferenceoftheA.P.is

[2011]

Options:

A.α −β

B. α−β

100

C.β −α

D. α−β

200

Answer:B

Let'′a′isthefirsttermand'd'isthecommondifferenceof anA .P.

Now,Accordingtothequestion100a100 =50a50

100(a+99d ) = 50(a+49d )

⇒2a +198d =a+49d ⇒a+149d =0

Hence,T150 =a+149d =0

nthtermofthegivenseries

=Tn= (n−1)2+ (n−1)n+n2

=((n−1)3−n3)

(n−1) − n=n3− (n−1)3

⇒Sn=

∑

k=1

[k3− (k−1)3] ⇒ 8000 =n3

⇒n=20whichisanaturalnumber.

Hence,boththegivenstatementsaretrue.

andstatement2iscorrectexplanationforstatement1.

Solution:

-------------------------------------------------------------------------------------------------

Question256

Ifthesumoftheseries12+2.22+32+2.42+52+ .. . 2.62+ ...uptonterms,

whenniseven,is n(n+1)2

2,thenthesumoftheseries,whennisodd,is

[OnlineMay26,2012]

Options:

A.n2(n+1)

B. n2(n−1)

C. n2(n+1)

D.n2(n−1)

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question257

Amansaves₹200ineachofthefirstthreemonthsofhisservice.Ineachof

thesubsequentmonthshissavingincreasesby₹40morethanthesavingof

immediatelypreviousmonth.Histotalsavingfromthestartofservicewillbe

₹11040after

[2011]

Options:

A.19months

B.20months

C.21months

D.18months

LetA.P.bea,a+d,a+2d , .........

a2+a4+ .......... + a200 =α

⇒100

2[2(a+d) + (100 −1)2d ] = α.. . (i)

anda1+a3+a5+ ........ + a199 =β

⇒100

2[2a + (100 −1)2d ] = β...(ii)

Subtracting(ii)from(i),weget

d=α−β

100

Ifnisodd,therequiredsumis

12+2.22+32+2.42+ ..... + 2(n−1)2+n2

=(n−1)(n−1+1)2

2+n2( ∵n−1iseven)

=n−1

2+1 n2=n2(n+1)

( )

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question258

Apersonistocount4500currencynotes.Letandenotethenumberofnotes

hecountsinthen th minute.Ifa1=a2= ... = a10 =150anda10,a11, ...arein

anAPwithcommondifference-2,thenthetimetakenbyhimtocountall

notesis

[2010]

Options:

A.34minutes

B.125minutes

C.135minutes

D.24minutes

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question259

Thesumtoinfinitetermoftheseries1 +2

3+6

32+10

33+14

34+ ⋅sis

[2009]

Options:

A.3

B.4

C.6

D.2

Letnumberofmonths = n

∴200 ×3+ ( 240 +280 +320 + ... + (n−3)th term) = 11040

⇒n−3

2[2×240 + (n−4) × 40] = 11040 −600

⇒(n−3)[240 +20n −80] = 10440

⇒(n−3)(20n +160) = 10440

⇒ (n−3)(n+8) = 522

⇒n2+5n −546 =0

⇒(n+26)(n−21) = 0

∴n=21

Till10 th minutenumberofcountednotes = 1500

Remainingnotes = 4500 −1500 =3000

3000 =n

2[2×148 + (n−1)(−2)] = n[148 −n+1]

n2−149n +3000 =0

⇒n=125,24

Butn=125isnotpossible

∴Totaltime = 24 +10 =34minutes.

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question260

Thefirsttwotermsofageometricprogressionaddupto12.thesumofthe

thirdandthefourthtermsis48.Ifthetermsofthegeometricprogression

arealternatelypositiveandnegative,thenthefirsttermis

[2008]

Options:

A.-4

B.-12

C.12

D.4

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question261

Inageometricprogressionconsistingofpositiveterms,eachtermequalsthe

sumofthenexttwoterms.Thenthecommonratioofitsprogressionis

equals

[2007]

Options:

A.√5

LetS=1+2

3+6

32+10

33+14

34+ ..... . ∞...(i)

Multiplyingbothsidesby 1

3,weget

3S=1

3+2

32+6

33+10

34+ ....... . ∞...(ii)

Subtractingeqn.(ii)fromeqn.(i),weget

3S=1+1

3+4

32+4

33+4

34+ ........ . ∞

⇒2

3S=4

3+4

32+4

33+4

34+ ........ . ∞

⇒2

3S=

1−1

3×3

2⇒S=3

AT Q

a+ar =12

ar2+ar3=48

⇒ar2(1+r)

a(1+r)=48

12 ⇒r2=4, ⇒r= −2

( ∵termsarealternately+veand−ve )

⇒a= −12

B.1

2(√5−1)

C. 1

2(1− √5)

D.1

2√5

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question262

Thesumofseries 1

2!−1

3!+1

4!− ......uptoinfinityis

[2007]

Options:

A.e

−1

B.e

C.e−2

D.e−1

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question263

Leta1,a2,a3............betermsonA.P.If a1+a2+ ......... . ap

a1+a2+ ........... + aq

=p2

q2,p≠q,then a6

a21

equals

[2006]

Options:

A. 41

Lettheseriesa,ar,ar2, ....areingeometricprogression.

Giventhat,a=ar +ar2

⇒1=r+r2⇒r2+r−1=0

⇒r=−1± √1−4×−1

2=−1± √5

2

⇒r=√5−1

2[ ∵termsofG.P.arepositive∴rshouldbepositive]

Weknowthatex=1+x+x2

2!+x3

3!+ ........ . ∞

Putx = −1

∴e−1=1−1+1

2!−1

3!+1

4!...... . ∞

∴e−1=1

2!−1

3!+1

4!−1

5!........ . ∞

B. 7

C. 2

D. 11

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question264

Thevalueof 10

∑

k=1sin 2kπ

11 +i cos 2kπ

11 is

[2006]

Options:

A.i

B.1

C.-1

D.−i

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question265

( )

Giventhat

2[2a1+ (p−1)d]

2[2a1+ (q−1)d]

=p2

q2

⇒2a1+ (p−1)d

2a1+ (q−1)d=p

q

Putp=11andq=41

a1+5d

a1+20d =11

41 ⇒a6

a21

=11

∑

k=1

sin 2kπ

11 +i cos 2kπ

11 

∑

k=1

cos 2kπ

11 −i sin 2kπ

11 [∵eiθ =cos θ +i sin θ]

∑

k=1

−2kπ

11 i=i

∑

k=0

−2kπ

11 i−1

=i 1 +e

−2π

11i+e

−4π

11i+ .. . .11terms] − i

=i1−e

−2π

1−e

−2π

11i

−i=i1−e−2πi

1−e

−2π

11i

−i

=i×0−i[∵e−2πi =1]

= −i

( )

{ }

[

[( ) ][ ]

Iftheexpansioninpowersofxofthefunction 1

(1−ax)(1−bx)is

a0+a1x+a2x2+a3x3......thenanis

[2006]

Options:

A. bn−an

b−a

B. an−bn

b−a

C. an+1−bn+1

b−a

D. bn+1−an+1

b−a

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question266

Ifa1,a2, ........., anareinH.P.,thentheexpression

a1a2+a2a3+ .......... + an−1anisequalto

[2006]

Options:

A.n(a1−an)

B.(n−1)(a1−an)

C.na1an

D.(n−1)a1an

Answer:D

Solution:

(1−ax)−1(1−bx)−1

= (1+ax +a2x2+ ...)(1+bx +b2x2+ ...)

∴Coefficientofxn

xn=bn+abn−1+a2bn−2+ ...... + an−1b+an

{whichisaG.P.withr=a

b

Itssumis =

bn1−a

n+1

1−a



=bn+1−an+1

b−a∴an=bn+1−an+1

b−a

[ ( ) ] }

∵a1,a2,a3..... . anareinH.P.

∴1

... 1

areinA.P.

∴1

−1

= ......... = 1

−1

an−1

=d(say)

-------------------------------------------------------------------------------------------------

Question267

Ifthecoefficientsofrth,(r+1)th,and(r+2)thtermsinthethebinomial

expansionof(1+y)mareinA.P.,thenmandrsatisfytheequation

[2005]

Options:

A.m2−m(4r −1) + 4r2−2=0

B.m2−m(4r +1) + 4r2+2=0

C.m2−m(4r +1) + 4r2−2=0

D.m2−m(4r −1) + 4r2+2=0

Answer:C

Solution:

-------------------------------------------------------------------------------------------------

Question268

Ifx =∞

∑

n=0an,y=∞

∑

n=0bn,z=∞

∑

n=0cnwherea,b,careinA.Pand

|a| < 1, | b| <1, | c| <1thenx,y,zarein

[2005]

Options:

A.G.P.

B.A.P.

C.Arithmetic-GeometricProgression

D.H.P.

Thena1a2=a1−a2

d,a2a3=a2−a3

d,

.......... . an−1an=an−1−an

d

Addingallequations,weget

∴a1a2+a2a3+ ......... + an−1an

=a1−a2

d+a2−a3

d+ .... + an−1−an

d

d[a1−a2+a2−a3+ .... + an−1−an] = a1−an

d

Also, 1

+ (n−1)d

⇒a1−an

a1an

= (n−1)d

⇒a1−an

d= (n−1)a1an

Whichistherequiredresult.

Coeficientofrth , (r+1)th and(r+2)th termsismCr−1,mCrandmCr+1resp.

GiventhatmCr−1,mCr,mCr+1areinA.P.

2mCr=mCr−1+mCr+1

⇒2=

mCr−1

mCr

mCr+1

mCr

m−r+1+m−r

r+1

⇒m2−m(4r +1) + 4r2−2=0

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question269

Thesumoftheseries1 +1

4.2!+1

16.4!+1

64.6!+ ....................adinf.is

[2005]

Options:

A. e−1

√e

B. e+1

√e

C. e−1

2√e

D. e+1

2√e

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question270

LetT rbetherthtermofanA.P.whosefirsttermisaandcommondifference

isd .Ifforsomepositiveintegersm,n,m≠n,Tm=1

nandT n=1

m,thena −d

equals

[2004]

Options:

∞

∑

n=0

an=1

1−a⇒a=1−1

x

∞

∑

n=0

bn=1

1−b⇒b=1−1

y

∞

∑

n=0

cn=1

1−c⇒c=1−1

z

a,b,careinA.P.⇒ 2b =a+c

2 1 −1

y=1−1

x+1−1

z

y=1

x+1

z⇒x,y,zareinH.P.

( )

Weknowthat

ex+e−x

2=1+x2

2!+x4

4!+x6

6!............

Puttingx=1

2,weget

1+1

4.2!+1

16.4!+1

64.6!+ ..... . ∞=e

2+e

−1

2

√e+1

√e

2=e+1

2√e

A. 1

m+1

B.1

C. 1

D.0

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question271

Lettwonumbershavearithmeticmean9andgeometricmean4.Thenthese

numbersaretherootsofthequadraticequation

[2004]

Options:

A.x2−18x −16 =0

B.x2−18x +16 =0

C.x2+18x −16 =0

D.x2+18x +16 =0

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question272

Thesumofseries 1

2!+1

4!+1

6!+ .....is

[2004]

Options:

A. (e2−2)

B. (e−1)2

Tm=a+ (m−1)d=1

n...(i)

Tn=a+ (n−1)d=1

m...(ii)

Subtracting(ii)from(i),weget

(m−n)d=1

n−1

m⇒d=1

mn

From(i)a=1

mn ⇒a−d=0

Lettwonumbersbeaandbthen a+b

2=9

⇒a+b=18and√ab =4⇒ab =16

∴Equationwithrootsaandbis

x2− (a+b)x+ab =0⇒x2−18x +16 =0

C. (e2−1)

D. (e2−1)

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question273

Thesumofthefirstntermsoftheseries12+2.22+32+2.42+52+2.62+ ...

is n(n+1)2

2whenniseven.Whennisoddthesum

[2004]

Options:

A. n(n+1)

B. n2(n+1)

C. n(n+1)2

D. 3n(n+1)

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question274

IfSn=n

∑

r=0

nCr

andtn=n

∑

r=0

nCr

,then tn

isequalto

[2004]

[ ]

Weknowthat

ex=1+x

1!+x2

2!+x3

3!... . ∞

∴e=1+1

1!+1

2!+1

3!+ ......

ande−1=1−1

1!+1

2!−1

3!+ .......

∴e+e−1=2 1 +1

2!+1

4!+ ... 

∴1

2!+1

4!+1

6!+ ...... = e+e−1

2−1

=e2+1−2e

2e =(e−1)2

[ ]

Ifnisodd,therequiredsumis

12+2.22+32+2.42+ ...... + 2⋅ (n−1)2+n2

=(n−1)(n−1+1)2

2+n2

[ ∵(n−1)iseven

∴usinggivenformulaforthesumof(n−1)terms.

=n−1

2+1 n2=n2(n+1)

( )

Options:

A. 2n −1

B. 1

2n−1

C.n −1

D. 1

Answer:D

Solution:

-------------------------------------------------------------------------------------------------

Question275

Ifthesumoftherootsofthequadraticequationax2+bx +c=0isequalto

thesumofthesquaresoftheirreciprocals,then a

c,b

aand c

barein

[2003]

Options:

A.Arithmetic-GeometricProgression

B.ArithmeticProgression

C.GeometricProgression

D.HarmonicProgression.

Answer:D

Solution:

Sn=1

nC0

nC1

nC2

+ .... + 1

nCn



tn=0

nC0

nC1

nC2

+ .... + n

nCn



tn=n

nCn

+n−1

nCn−1

+n−2

nCn−2

+ ... + 0

nC0



Adding(i)and(ii),weget,

2tn= (n)1

nC0

nC1

+ ... 1

nCn

=nSn

∵nCr=nCn−r

∴tn

[ ]

ax2+bx +c=0,α+β=−b

a,αβ =c

a

AT Q,α+β=1

α2+1

β2

α+β=α2+β2

α2β2⇒ − b

a2−2c



Onsimplification2a2c=ab2+bc2

⇒2a

b=c

a+b

cDividebothsidebyabc ]

⇒c

a,a

b,b

careinA.P.

[

-------------------------------------------------------------------------------------------------

Question276

Thesumoftheseries 1

1.2 −1

2.3 +1

3.4 ............upto∞isequalto

[2003]

Options:

A.loge

B.2loge2

C.loge2−1

D.loge2

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question277

If1,log9(31−x+2), log3(4.3x−1)areinA.P.thenxequals

[2002]

Options:

A.log34

B.1 −log34

C.1 −log43

D.log43

Answer:B

Solution:

( )

∴a

c,b

a, & c

bareinH.P.

LetS=1

1.2 −1

2.3 +1

3.4.............. . ∞

Tn=1

n(n+1)=1

n−1

n+1

∴S=1

1−1

2−1

3+1

3−1

4−1

5....

=1−21

2−1

3+1

4−1

5.............. . ∞

∵log(1+x) = x−x2

2+x3

3−x4

4+ ..... . ∞

=1−2[−log(1+1) + 1] = 2 log 2 −1=log 4

( )

( ) ( ) ( ) ( )

[ ]

( )

1,log9(31−x+2), log3(4.3x−1)areinA.P.

∵a,b,careinA.Pthenb=a+c

⇒2log9(31−x+2) = 1+log3(4.3x−1)

-------------------------------------------------------------------------------------------------

Question278

SumofinfinitenumberoftermsofGPis20andsumoftheirsquareis100.

ThecommonratioofGPis

[2002]

Options:

A.5

B.3

C.8

D.1

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question279

FifthtermofaGPis2,thentheproductofits9termsis

[2002]

Options:

A.256

B.512

C.1024

∵logbqap=p

qlogba

⇒log3(31−x+2) = log33+log3(4.3x−1)

⇒log3(31−x+2) = log3[3(4⋅3x−1)]

⇒31−x+2=3(4.3x−1)

⇒3.3−x+2=12.3x−3Put3x=t

⇒3

t+2=12t −3⇒12t2−5t −3=0

Hencet= − 1

3,3

4

⇒3x=3

4as3x≠ −ve )

⇒x=log3

4orx=log33−log34

⇒x=1−log34

(

( )

Leta= firsttermofG.P.andr= commonratioofG.P.

ThenG.P.isa,ar,ar2

GivenS∞=20 ⇒a

1−r=20

⇒a=20(1−r)...(i)

Alsoa2+a2r2+a2r4+ ...to∞=100

⇒a2

1−r2=100 ⇒[20(1−r)]2

1−r2=100 [from(i) ]

⇒400(1−r)2

(1−r)(1+r)=100 ⇒4(1−r) = 1+r

⇒1+r=4−4r ⇒5r =3⇒r=3∕5

D.noneofthese

Answer:B

Solution:

-------------------------------------------------------------------------------------------------

Question280

13−23+33−43+ ... + 93=

[2002]

Options:

A.425

B.-425

C.475

D.-475

Answer:A

Solution:

-------------------------------------------------------------------------------------------------

Question281

Thevalueof21∕4⋅41∕8⋅81∕16...∞is

[2002]

Options:

A.1

B.2

C.3

D.4

Answer:B

Solution:

∵a4=2⇒ar4=2

Now,a×ar ×ar2×ar3×ar4×ar5×ar6×ar7×ar8

=a9r36 = (ar4)9=29=512

13−23+33−43+ ....... + 93

=13+23+33+ ...... + 93−2(23+43+63+83)

∵Σn3=n(n+1)

2

=9×10

2−2.23[13+23+33+43]

= (45)2−16 ⋅4×5

2=2025 −1600 =425

[ ( ) ]

[ ]

-------------------------------------------------------------------------------------------------

Now,letS=1

4+2

8+3

16 + ........ . ∞..... . (i)

2S=1

8+2

16 + ....... . ∞.....(ii)

Subtracting(ii)from(i)

⇒1

2S=1

4+1

8+1

16 + ........ . ∞

or 1

2S=a

1−r=1∕4

1−1∕2=1

2⇒S=1

∴P=2S=2

LetP=21∕4⋅22∕8⋅23∕16........ . ∞

=21∕4+2∕8+3∕16 + ........ . ∞